Linear Algebra [4 ed.] 9780136800309
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A.1 Euclidean spaces Section 1.1 1. Write a three-column zero vector, and a three-column nonzero vector, and a five-column zero vector and a five-column nonzero vector. 2. Suppose that the buyer for a manufacturing plant must order different quantities of oil, paper, steel, and plastics. He will order 40 units of oil, 50 units of paper, 80 units of steel, and 20 units of plastics. Write the quantities in a single vector. 3. Suppose that a student’s course marks for quiz 1, quiz 2, test 1, test 2, and the final exam are 70, 85, 80, 75, and 90, respectively. Write his marks as a column vector. 4. Let
x − 2y ⎛ ⎞ → a = ⎜ 2x − y ⎟ ⎝
5.
|x| → a = ( ) y
6.
2
b
⎝
and
→ b
= (
1
2
⎞
= ⎜ −2 ⎟ .
⎠
2z
x−y → a = ( )
and
⎛
→
).
1
→ a =
Find all x, y, z ∈ R that
→ b .
⎠
Find all x, y ∈ R such that
→ a =
→ b .
4
and
→ b
= (
2
).
Find all x, y ∈ R such that
→ a −
→ b
is a nonzero vector.
x+y
4
7. Let P (1, x2 ), Q(3, 4), P1 (4, 5), and Q1 (6, 1). Find all possible values of x ∈ R such that − − →
− − − →
P Q = P 1 Q1 .
8. A company with 553 employees lists each employee’s salary as a component of a vector If a 6% salary increase has been approved, find the vector involving 9. Let …
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110 ⎛ ⎞ → a = ⎜ 88 ⎟ ⎝
⎠
→ a
→ a
in R553 .
that gives all the new salaries.
denote the current prices of three items at a store. Suppose that the store
Pearson eText ⎝
40
⎠
announces a sale so that the price of each item is reduced by 20%. a. Find a three-vector that gives the price changes for the three items. b. Find a three-vector that gives the new prices of the three items.
10. Let
⎛ → a = ⎜
−1
⎝
2 3
⎞
→
⎟,
b
2
⎛
⎛ → = ⎜ −2 ⎟ , c = ⎜
⎠
⎝
i.
→ → → 2 a − b +5 c ;
ii.
− → → → 4 a + αb − 2 c
0
9
⎛
⎞
⎠
⎞
⎝
⎛
3x
11. Find x, y, and z such that ⎜ 4y ⎟ + ⎜ ⎝
12. Let
→ u = (1,
− 1, 0, 2),
2z
⎠
⎝
8 −6
→ v = (−2,
a. Find a vector
→ 4 a ∈ R
b. Find a vector
→ a
⎞
0 −1
⎞ ⎟ , α ∈ R.
Compute
⎠
⎛
0
⎞
⎟ = ⎜0⎟. ⎠
⎝
0
⎠
− 1, 1, − 4), →
such that 2 u
1
3
and
→ w = (3, 2,
− 1, 0).
→ → → −3 v − a = w .
→ → → → → → [2 u − 3 v + a ] = 2 w + u − 2 a .
2
13. Determine whether 1.
→ ∥ a ∥
→ a = (1, 2, 3)
→ b ,
and
2.
→ a = (−1, 0, 1, 2)
3.
→ a = (1,
→ b
= (−2,
and
− 1, 1, 2)
− 4,
− 6).
→
and
b
= (−3, 0, 3, 6).
→ b
= (−2, 2, 2, 3).
14. Let x, y, a, b ∈ R with x ≠ y. Let P (x, 2x), Q(y, 2y), P1 (a, 2a), and Q1 (b, 2b) be points in …
2/5
, y, 2
,
≠ y − − →
R .
∥
(
,
Pearson eText ), Q(y, y),
1(
,
),
Q1 ( ,
)
p
− − − →
Show that P Q ∥ P1 Q1 .
15. Let P (x, 0), Q(3, x2 ), P1 (2x, 1), and Q1 (6, x2 ). Find all possible values of x ∈ R such that − − → ∥− − − → P Q ∥ P 1 Q1 .
16. Let
⎛ → a = ⎜ ⎝
−1 2 3
⎞
→
⎟,
b
⎠
⎛
2
⎛ → = ⎜ −2 ⎟ , c = ⎜ ⎝
0
⎞
⎠
⎝
3 0 −1
⎞ ⎟,
and α ∈ R.
⎠
Compute a.
→ → a ⋅ b ;
b.
→ → a ⋅ c ;
c. d. 17. Let
→
→ b ⋅ c ;
→ → → a ⋅ ( b + c ).
→ → u = (1, − 1, 0, 2), v = (−2, − 1, 1, − 4), → → u ⋅ v ,
→ → u ⋅ w ,
and
→ w = (3, 2, − 1,
0).
→ → w ⋅ v .
Compute and 18. Assume that the percentages for homework, test 1, test 2, and the final exam for a course are 10%, 25%, 25%, and 40%, respectively. The total marks for homework, test 1, test 2, and the final exam are 10, 50, 50, and 90, respectively. A student’s corresponding marks are 8, 46, 48, and 81, respectively. What is the student’s final mark out of 100? 19. Assume that the percentages for homework, test 1, test 2, and the final exam for a course are 14%, 20%, 20%, and 46%, respectively. The total marks for homework, test 1, test 2, and the final exam are 14, 60, 60, and 80, respectively. A student’s corresponding marks are 12, 45, 51, and 60, respectively. What is the student’s final mark out of 100? 20. A manufacturer produces three different types of products. The demand for the products is denoted by the vector
→ a = (10, 20, 30).
→
The price per unit for the products is given by the vector
If the demand is met, how much money will the manufacturer receive? 21. A company pays four groups of employees a salary. The numbers of the employees for the four b
…
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= ($200, $150, $100).
Pearson eText
groups are expressed by a vector
→ a = (5, 20, 40).
The payments for the groups are expressed by
→
a vector b = ($100, 000, $80, 000, $60, 000). Use the dot product to calculate the total amount of money the company paid its employees. 22. There are three students who may buy Calculus or algebra books. Use the dot product to find the total number of students who buy both calculus and algebra books. 23. There are n students who may buy a calculus or algebra books (n ≥ 2). Use the dot product to find the total number of students who buy both calculus and algebra books. 24. Assume that a person A has contracted a contagious disease and has direct contacts with four →
people: P1, P2, P3, and P4. We denote the contacts by a vector a := (a11 , a12 , a13 , a14 ), where if the person A has made contact with the person Pj, then a1j = 1, and if the person A has made no contact with the person Pj, then a1j = 0. Now we suppose that the four people then have had a variety of direct contacts with another individual B, which we denote by a vector → b bj1
where if the person Pj has made contact with the person B, then = 1, and if the person Pj has made no contact with the person B, then bj1 = 0. i. Find the total number of indirect contacts between A and B. := (b11 , b21 , b31 , b41 ),
→
ii. If a = (1, 0, 1, between A and B.
…
4/5
1)
and
→ a = (1, 1, 1, 1),
find the total number of indirect contacts
Pearson eText
…
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Pearson eText
Solution 1. 1. 2. 2. 3. 3. 4. 4.
(0, 0, 0)
T
, (1, 0, 1)
T
, (0, 0, 0, 0, 0)
(oil, paper, steel, plastics)
T
T
, (−1, 0, 1, 0, 1)
= (40, 50, 80, 20)
(quiz 1, quiz 2, test 1, test 2, final exam) → a =
→ b
T
y = 2x + 2 = 2(−2) + 2 = −2
5. 5.
1 2
→ a =
Hence,
→ , a = → b
.
.
= (70, 85, 80, 75, 90)
T
.
if and only if x − 2y = 2
(1)
2x − y = −2
(2)
2z = 1.
(3)
(1) − (2) × 2 : (x − 2y) − 2(2x − y) = 2 − 2(−2)
and z =
T
T
and by (3), we have z =
1 2
.
Therefore, when x = −2, y = −2
→ b .
if and only if |x| = 1 and y 2
→ → a = b
and x = −2. By (2), we get
if x = 1 and y = 2;
= 4.
x = 1
Hence, x = 1 or x = −1 and y = 2 or y = −2.
and y = −2;
x = −1
and y = 2; or x = −1 and
y = −2.
6. 6.
We first find all x, y such that → a −
→ → → a − b = 0 .
→ b
= (
x−y 4
Because
)−(
2 x+y
) = (
x−y−2 4−x−y
) = (
0
),
0
we obtain https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
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Pearson eText
→
Hence when x = 3 and y = 1, a When x ≠ 3 or y ≠ 1, 7. 7.
b
→ → → a − b ≠ 0 ,
− − →
Because P Q
−
→ =
(1)
4 − x − y = 0.
(2)
and y = 1. By (1), we get x = y + 2 = 1 + 2 = 3.
(1) + (2) : (x − y − 2) + (4 − x − y) = 0 →
x−y−2 = 0
Now, we can find all x, y such that
0 .
that is,
→ → a − b
2
2
= (3 − 1, 4 − x ) = (2, 4 − x )
→ → → a − b ≠ 0 .
is a nonzero vector. and
− − − → P 1 Q1 = (6 − 4, 1 − 5) = (2, − 4),
− − →
then P Q
− − − →
= P 1 Q1 – x = 2√2. → 1.06 a .
if and only if 4 − x2
= −4
if and only if x2
= 8
–
if and only if x = −2√2 or
8. 8. 9. 9.
a.
−22 ⎛ ⎞ → −0.2 a = ⎜ −17.6 ⎟ . ⎝
b.
−8
⎠
88 ⎛ ⎞ → 0.8 a = ⎜ 70.4 ⎟ . ⎝
32
⎠
10. 10. ⎛ → → → 2 a − b + 5 c = 2⎜ ⎝
i.
−1 2 3
⎞
⎛
2
⎞
⎛
⎟ − ⎜ −2 ⎟ + 5 ⎜ ⎠
⎝
0
⎠
⎝
3 0 −1
⎞ ⎟ ⎠
2 2 15 2 2 + 15 11 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
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⎛ = ⎜ ⎝
−2
⎞
⎠
6
2 3
ii. = ⎜ ⎝
8 12
⎞
2α
⎞
⎠
⎝
9 + 3x
⎞
0
⎛
⎠
⎝
⎛
⎠
⎝
2
0 −5
⎝
⎟ = ⎜ ⎠
⎞
6
⎞
3
⎟ = ⎜
0 −2
⎠
0
⎝
−1
−10 + 2α
⎛
8 − 2α
⎝
4+2+0
⎝
⎛
⎠
0
−2 − 2 + 15
⎛
⎟ + α ⎜ −2 ⎟ − 2 ⎜
⎛
⎠
0
0
⎞
⎟ − ⎜ −2α ⎟ − ⎜
⎛
11. 11.
⎛
⎝
−1
⎝
−4
⎛
Pearson eText 15 ⎞ ⎛ ⎞
⎟ − ⎜ −2 ⎟ + ⎜
4
⎛ → → → 4 a + α b − 2 c = 4⎜
⎛
2
6−0−5
⎞
⎛
11
⎞
⎟ = ⎜ 6 ⎟. ⎠
⎝
1
⎠
⎞ ⎟ ⎠
⎞ ⎟. ⎠
14
⎞
Because ⎜ 4y + 8 ⎟ = ⎜ 0 ⎟ , we have ⎝
2z − 6
⎠
⎝
0
⎠
9 + 3x = 0,
This implies x = −3, 12. 12. →
a. Because 2 u
4y + 8 = 0,
2z − 6 = 0.
y = −2, z = 3.
→ → → −3 v − a = w ,
we have
→ → → → a = 2 u −3 v − w
b. Because
1 2
=
2(1,
− 1, 0, 2) − 3(−2,
=
(2,
=
(2 + 6 − 3, − 2 + 3 − 2, 0 − 3 + 1, 4 + 12 − 0) = (5, − 1, − 2, 16).
− 2, 0, 4) − (−6,
− 1, 1,
− 3, 3,
→ → → → → → [2 u − 3 v + a ] = 2 w + u − 2 a ,
− 4) − (3, 2,
− 12) − (3, 2,
− 1, 0) − 1, 0)
we have
→ → → → → → → → → 2 u − 3 v + a = 2[2 w + u − 2 a ] = 4 w + 2 u − 4 a . https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
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2 u
3 v + a
Pearson eText 2[2 w + u
2 a ]
4w + 2 u
4 a .
Hence, → 5 a = =
→ → 3 v + 4 4 = 3(−2, − 1, 1, − 4) + 4(3, 2, − 1, 0) (−6, − 3, 3, − 12) + (12, 8, − 4, 0) = (6, 5, − 1, − 12)
and 1 6 1 12 → (6, 5, − 1, − 12) = ( , 1, − , − ). a = 5 5 5 5
13. 13. 1. Because 2. Because 3. Because
14. 14.
→ b → b → b
→ → → = −2 a , b || a . → = 3 a , → ≠ k a
→
→ b || a .
for any k ∈ R,
→ b
→ || a .
− − →
Because P Q
= (y − x, 2y − 2x) = (y − x)(1, 2), − − − → P 1 Q1 = (b − a, 2b − 2a) = (b − a)(1, 2)
and x ≠ y,
we have − − − → P 1 Q1 =
− → a−b − P Q. x−y
− − → − − − →
Hence, P Q||P1 Q1 . 15. 15.
− − →
Because P Q
2
= (3 − x, x
2
− 0) = (3 − x, x )
− − − →
and P1 Q1
2
= (6 − 2x, x
− 1),
then
https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
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Pearson eText − − → − − − → P Q||P 1 Q1
− − − →
if and only if there exists k ∈ R such that P1 Q1 2
(6 − 2x, x
− − → = kP Q,
that is,
2
2
− 1) = k(3 − x, x ) = (k(3 − x), kx ).
if and only if 6 − 2x = k(3 − x) and x2 − 1 = kx2 if and only if 3(2 − k) = (2 − k)x and 2 (1 − k)x = 1. These last two equations imply k ≠ 1 and k ≠ 2 because if k = 1 or k = 2, the last equation implies 0 = 1 or x2 = −1, respectively, and it is impossible. If k ≠ 1 and k ≠ 2, then x = 3
and k =
8 9
− − → − − − →
.
Hence, only when x = 3, P Q||P1 Q1 .
16. 16. a.
2 ⎛ ⎞ → → a ⋅ b = (−1, 2, 3) ⎜ −2 ⎟ = (−1)(2) + 2(−2) + (3)(0) = −6. ⎝
b.
⎛ → → a ⋅ c = (−1, 2, 3) ⎜ ⎝
c.
0 3 0 −1
⎠
⎞ ⎟ = (−1)(3) + 2(0) + (3)(−1) = −6. ⎠
→ b ⋅ c = (2,
3
⎛
→
− 2, 0) ⎜ ⎝
0 −1
⎞ ⎟ = (2)(3) + (−2)(0) + (0)(−1) = 6. ⎠
2 ⎛⎛ ⎞ ⎛ → → → a ⋅ ( b + c ) = (−1, 2, 3) ⎜⎜ −2 ⎟ + ⎜ ⎝⎝
d. ⎛
5
0
⎠
⎝
3 0 −1
⎞⎞ ⎟⎟ ⎠⎠
⎞
= (−1, 2, 3) ⎜ −2 ⎟ = (−1)(5) + 2(−2) + 3(−1) = −12. ⎝
−1
⎠
−2 ⎛ ⎞ https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
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⎛ → → u ⋅ v = (1,
⎞
Pearson eText
⎜ −1 ⎟ − 1, 0, 2) ⎜ ⎟ = 1(−2) + (−1)(−1) + 0(1) + 2(−4) ⎜ 1 ⎟ ⎝
−4
⎠
= −2 + 1 + 0 − 8 = −9. ⎛
3
⎞
→ → ⎜ 2 ⎟ u ⋅ w = (1, − 1, 0, 2) ⎜ ⎟ = 1(3) + (−1)2 + 0(−1) + 2(0) ⎜ −1 ⎟
17. 17.
⎝
0
⎠
= 3 − 2 + 0 + 0 = 1. ⎛ → → w ⋅ v = (3, 2,
−2
⎞
⎜ −1 ⎟ − 1, 0) ⎜ ⎟ = 3(−2) + 2(−1) + (−1)(1) + 0(−4) ⎜ 1 ⎟ ⎝
−4
⎠
= −6 − 2 − 1 + 0 = −9.
18. 18.
Let → a = (10/10, 25/50, 25/50, 40/90) = (1, 1/2, 1/2, 4/9)
and
→ b
= (8, 46, 48, 81).
→ a ⋅
Then the final mark out of 100 for the student is
→ b
= (1, 1/2, 1/2, 4/9) ⋅ (8, 46, 48, 81) = (1)(8) + (1/2)(46) + (1/2)(48) + (4/9)(81) = 8 + 23 + 24 + 36 = 91.
19. 19.
Let → a = (14/14, 20/60, 20/60, 46/80) = (1, 1/3, 1/3, 23/40)
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(
and
→ b
= (12, 45, 51, 60).
/
,
/
,
Pearson eText / , /
)
( ,
/ ,
/ ,
/
)
Then the final mark out of 100 for the student is
→ → a ⋅ b = (1, 1/3, 1/3, 23/40) ⋅ (12, 45, 51, 60) = (1)(12) + (1/3)(45) + (1/3)(51) + (23/40)(60) = 12 + 15 + 17 + 34.5 = 78.5
20. 20. 21. 21.
→
→
The total cash received is a ⋅ b = 10(200) + 20(150) + 30(100) = $8000. The total amount of money the company paid is the dot product → a ⋅
→ b
= 5(100, 000) + 20(80, 000) + 40(60, 000) = $4, 500, 000
because there are five employees with salary $100,000, 20 with $80,000, and 40 with $60,000. We denote the calculus book by book 1 and the algebra book by book 2. If student 22. 22. then we denote by aij = 1. If student i doesn’t i(i = 1, 2, 3) wants to buy book j(j = 1, 2), want to buy book j, then we denote by aij
= 0.
We write
→ a = (a11 , b21 , a31 )
and
→
Then for i = 1, 2, 3, if ai1 ai2 = 1, then student i wants to buy both calculus and algebra books, and if ai1 ai2 = 0, then student i doesn’t want to buy both calculus and algebra books. Hence, the total number of students who buy both calculus and algebra books is equal to b
= (a12 , b22 , a32 ).
→ a ⋅
23. 23.
We denote by aij
aij = 0
= 1
→ b
= a11 a12 + a21 a22 + a31 a32 .
if student i(i = 1, 2,
⋯ , n)
if student i doesn’t want to buy book j. We write
wants to buy book j(j = 1, 2) and by
→ a = (a11 , a21 , ⋯ , an1 )
and
→
Then the total number of students who buy both calculus and algebra books is equal to the dot product b
= (a12 , a22 , ⋯ , an2 ).
https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
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Pearson eText
→ a ⋅
→ b
= a11 a12 + a21 a21 + ⋯ + an1 an2 ).
24. 24. i. The total number of indirect contacts between A and B is given by → → a ⋅ b = a11 b11 + a12 b21 + a13 b31 + a14 b41 .
ii. The total number of indirect contacts between A and B is → → a ⋅ b = (1)(1) + (0)(1) + (1)(1) + (1)(1) = 3.
(A.1.1)
From the first term of (A.1.1), we see that A has contact with P1 and P1 has been in contact with B, so A and B have one indirect contact via P1, which is shown by (1)(1) = 1. Similarly, A and B have one indirect contact via each of P3 and P4. From the second term of (A.1.1), we see that A has no contact with P2 and P2 has been in contact with B, so A and B have no indirect contact, which is shown by (0)(1) = 0.
Section 1.2 1. For each of the following vectors, find its norm and normalize it to a unit vector. i. ii. iii. 2. Let
→ a = (1, 0, − 2); → b
= (1, 1, − 2);
→ c = (2, 2, − 2).
→ → n a , b ∈ R .
i.
Show that the following identities hold.
2 →∥ 2 →∥ 2 ∥→ ∥→ → 2 ∥→∥ ∥ ∥ ∥ a + b ∥ + ∥ a − b ∥ = 2∥ a ∥ + 2∥ b ∥ .
→ →∥ 2 →∥ 2 ∥→ ∥→ → https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
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ii.
→∥ ∥→ ∥ a + b ∥
→∥ ∥→ −∥ a − b ∥
→ = 4( a ⋅
Pearson eText
→
b ).
3. Evaluate the determinant of each of the following matrices. A = (
1
1
4
−2
) B = (
−2
2
1
−3
) C = (
2
−1
3
0
) D = (
−2
1
−6
3
)
4. For each pair of vectors, verify that Cauchy-Schwarz and triangle inequalities hold. i. ii. iii.
→ a = (1, 2),
→ b
= (2, − 1)
→ → a = (1, 2, 1), b = (0, 2, − 2) → a = (1,
→ − 1, 0, 1),
b
= (0,
− 1, 2, 1)
→
→ a = (2, 1)
to find x ∈ R such that 5. Let and b = (1, x). Use Theorem 1.2.3 6. For each pair of points P1 and P2 , find the distance between the two points. 1. P1 (1, − 2), P2 (−3, − 1) 2. P1 (3, − 1), P2 (−2, 1) 3. P1 (1, − 1, 5), P2 (1, − 3, 1) 4. P1 (0, − 1, 2), P2 (1, − 3, 1) 5. P1 (−1, − 1, 5, 3), P2 (0, − 2, − 3, − 1)
→ ∥ a ∥
→ b .
7. Let A(−1, − 2), B(2, 3), C(−1, 1), and D(0, 1) be four different points in R2 . Find the distance between the midpoints of the segments AB and CD. 8. For each pair of vectors, determine whether they are orthogonal. i. ii. iii.
→
→ a = (1, 2),
b
→ a = (1, 1, 1), → a = (1,
= (2,
− 1)
→ b
= (0, 2,
− 2)
→ − 1,
− 1),
b
= (1, 2,
− 2)
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iv.
9/12
iv. v.
a
(1,
→ a = (2,
1, 1),
b
(0, 2,
Pearson eText 1)
→ − 1, 0, 1),
b
= (0,
− 1, 3,
− 1)
9. For each pair of vectors, find all values of x ∈ R such that they are orthogonal. i. ii. iii.
→
→ a = (−1, 2),
2
b
= (2, x )
→ → a = (1, 1, 1), b = (x, 2, − 2) → → a = (5, x, 0, 1), b = (0, − 1, 3, − 1)
10. For each pair of vectors, find cos θ, where θ is the angle between them. 1. 2. 11. Let
→ a = (1, − 1, 1) → a = (1, 0, 1)
→ a = (6, 8)
i. ii.
→ a ⊥
and
and
and
→ b
= (1, 1, − 2);
→ b
= (−1,
→ b
= (1, x).
− 1,
− 1).
Find x ∈ R such that
→ b
→ ∥ → a ∥ b
iii. the angle between
→ a
and
→ b
is
π 4
.
12. Three vertices of a triangle are P 1 (6, 6, 5, 8), P 2 (6, 8, 6, 5), and P 3 (5, 7, 4, 6).
a. Show that the triangle is a right triangle. b. Find the area of the triangle. 13. Suppose three vertices of a triangle in R5 are P (4, 5, 5, 6, 7), O(3, 6, 4, 8, 4), Q(5, 7, 7, 13, 6).
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Pearson eText
a. Show that the triangle is a right triangle. b. Find the area of the triangle. 14. Four vertices of a quadrilateral are A(4, 5, 5, 7), B(6, 3, 3, 9), C(5, 2, 2, 8), D(3, 4, 4, 6).
a. Show that the quadrilateral is a rectangle. b. Find the area of the quadrilateral.
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Pearson eText
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Solution
Solution 1. 1. i.
− − − − − − − − − − − − − → – 2 2 2 || a || = √1 + 0 + (−2) = √5.
→ a → || a ||
ii.
→
− − − − − − − − − − − − −
2 2 || b || = √1 + 1 + (−2)
2
1
1 2 – (1, 0, − 2) = ( – , 0, − –). √5 √5 √5
=
– = √6.
→ b →
1 =
|| b ||
1
–, √6
–, √6
1
1
2 −
–). √6
− − − − − − − − − − − − − → − − – 2 2 2 √ || c || = = √12 = 2√3. 2 + 2 + (−2)
→ c → || c ||
2. 2.
− 2) = (
1 =
– 2√3
t
iii.
1
– (1, 1, √6
(
–, √3
–, √3
1 −
–). √3
By Theorem 1.2.1 (iii)and (iv), → || a +
→
2
b ||
→ + || a −
→ 2 → = (|| a || + 2( a ⋅
→
2
b ||
→
→ → → → 2 → 2 2 b ) + || b || ) + (|| a || − 2( a ⋅ b ) + || b || )
→ → 2 2 = 2|| a || + 2|| b || .
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Solution → || a +
→
2
b ||
→ − || a −
→ 2 → = (|| a || + 2( a ⋅
→
→
2
b ||
→
→ 2 → b ) + || b || ) − (|| a || − 2( a ⋅ 2
→
→
2
b ) + || b || )
→ → = 4( a ⋅ b ).
3. 3. ∣1 |A| = ∣ ∣4
∣ ∣ = −2 − 4 = −6. −2 ∣
∣ −2 |B| = ∣ ∣ 1
∣ ∣ = 6 − 2 = 4. −3 ∣
∣2 |C| = ∣ ∣3
−1 ∣ ∣ = 0 − (−3) = 3. 0 ∣
∣ −2 |D| = ∣ ∣ −6
1∣ ∣ = −6 − (−6) = 0. 3∣
1
2
4. 4. →
i. By computation, || a
− − − − − − − − − → − − − − − − – – 2 2 2 2 || = √1 + 2 = √5, || b || = √2 + (−1) = √5
and
→ → a ⋅ b = (1)(2) + (2)(−1) = 0.
Hence, → → → → – – | a ⋅ b | = 0 ≤ √5√5 = || a |||| b ||
and the Cauchy-Schwarz inequality holds. By computation, → → → → − − – – || a + b || = √10 ≤ √5 + √5 = || a || + || b ||
and the triangle inequality holds. →
ii. By computation, || a
→ – – || = √6, || b || = 2√2,
and
→ → a ⋅ b = 2.
Hence,
→ → → → – – | a ⋅ b | = 2 ≤ √6(2√2) = || a |||| b ||
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Solution
and the Cauchy-Schwarz inequality holds. By computation, → → → → – – – || a + b || = 3√2 ≤ √6 + 2√2 = || a || + || b ||
and the triangle inequality holds. →
iii. By computation, || a
→ – – || = √3, || b || = √6,
→ | a ⋅
and
→ → a ⋅ b = 2.
Hence,
→
→ → – – b | = 2 ≤ √3√6 = || a |||| b ||
and the Cauchy-Schwarz inequality holds. By computation, → → → → − − – – || a + b || = √13 ≤ √3 + √6 = || a || + || b ||
and the triangle inequality holds. 5. 5.
→
By computation, | a
→ ⋅
− − − − − − → – 2 2 b | = |2 + x|, || a || = √2 + 1 = √5,
and
→ − − − − − − − − − − − 2 2 2 || b || = √ 1 + x = √ 1 + x .
If
→ | a ⋅
→
→ → b | = || a |||| b ||,
then − − − − − – 2 |2 + x| = √5√ 1 + x
and (2 + x)
2
2
= 5(1 + x ).
This implies 2
4 + 4x + x
Hence, 2x − 1 = 0 and x =
1 2
2
2
= 5 + 5x and 4x
− 4x + 1 = (2x − 1)
2
= 0.
.
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Solution
6. 6. 1. 2.
− −−−− −−−− −−−− −−−− −− −
− − − →
||P 1 P 2 || = √(−3 − 1) ||P 1 P 2 || = √(−2 − 3) ||P 1 P 2 || = √(1 − 1) 2
+ (1 − (−1))
2
− − − − − − − − 2 2 = √4 + 1 = √17. − − − − − − − − −
2
+ (−3 − (−1))
+ (−4)
= √(−5)
2
2
+2
− − = √29.
2
2
+ (1 − 5)
2
– = 2√5.
− −−− −−− −−−− −−− −−−− −−− −−−− − −
||P 1 P 2 || = √(1 − 0)
2
+ (−3 − (−1))
− −−− −−− −−− −−− − − 2 = √1 + (−2)
2
+ (−1)
2
2
+ (1 − 2)
2
– = √6
− −−−− −−−− −−−− −−−− −−−− −−−− −−−− −−−− −−−− −
− − − →
||P 1 P 2 || = √(0 − 1)
2
+ (−2 − (−1))
− − −− −− −− −− −− −− −− − −− −− − 2 = √1 + (−1)
7. 7.
2
− −−− −−− −−− −−− − −
− − − →
5.
2
− −−− −−− −−−− −−− −−−− −−− −−−− − −
2 = √0 + (−2)
4.
+ (−1 − (−2))
− − −− −− −− −− −− −− −− −− −
− − − →
− − − →
3.
2
2
+ (−8)
2
+ (−4)
2
2
+ (−3 − 5)
2
+ (−1 − 3)
2
− − − = √82.
By (1.2.14), two midpoints of the segments AB and CD are (
−1 + 2
,
−2 + 3
2
) = (
2
1
,
2
1
) and (
2
−1 + 0
,
1+1
2
2
) = (−
1
, 1) .
2
By (1.2.10) with n = 2, we obtain the distance between two midpoints − −− −−− −−− −−− −−− −− − − √(
1 2
+
1 2
2
)
+ (
1
2
− 1)
5 =
2
. 2
8. 8. i. ii.
→ a ⊥ → a ⊥
→ b → b
because because
→ a ⋅ → a ⋅
→ b
= (1)(2) + (2)(−1) = 0.
→ b
= (1)(0) + (1)(2) + (1)(−2) = 0.
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4/10
Solution
iii.
→ → a ⊥ b
because
→ → a ⋅ b = (1)(1) + (−1)(2) + (−1)(−2) = 1 ≠ 0
iv.
→ → a ⊥ b
because
→ → a ⋅ b = (1)(0) + (−1)(2) + (1)(−1) = −3 ≠ 0.
v.
→ → a ⊥ b
because
→ → a ⋅ b = (2)(0) + (−1)(−1) + (0)(3) + (1)(−1) = 0.
9. 9. i. If
→ → 2 2 a ⋅ b = (−1)(2) + (2)(x ) = −2 + 2x = 0,
x = −1, → →
ii. If
a ⋅
b
→ a ⊥
then x2
= 1.
Hence, when x = 1 or
→ b .
= (1)(x) + (1)(2) + (1)(−2) = x = 0,
then x = 0. Hence, when
→ → x = 0, a ⊥ b . → →
iii. If
a ⋅
b
= (5)(0) + (x)(−1) + (0)(3) + (1)(−1) = −x − 1 = 0, →
when x = −1, a
then x = −1. Hence,
→ ⊥
b .
10. 10. → → a ⋅ b = (1)(1) + (−1)(1) + (1)(−2) = −2, − − − − − − − − − − − − − − − − −− − − −− − −− − − → → – – 2 2 2 2 2 2 √ √ √ || a || = 1 + (−1) + 1 = 3 || b || = 1 + (1) + (−2) = √6.
1. By computation,
and
By Theorem 1.2.15, we have → a ⋅ cos θ =
2. By computation,
→ b
→ → || a |||| b ||
−2 =
2
– = − – – = − √3√6 3√2
– √2 . 3
→ → a ⋅ b = (1)(−1) + (0)(−1) + (1)(−1) = −2,
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Solution − − −− −− − −− − −
→ – 2 2 2 || a || = √ 1 + (0) + 1 = √2 and − −− −− −− −− −− −− −− −− − → – 2 2 2 || b || = √ (−1) + (−1) + (−1) = √3.
By Theorem 1.2.15,
cos θ =
→ → a ⋅ b → → || a |||| b ||
=
−2
2
– – = − – = − √2√3 √6
– √6
.
3
11. 11. i. If ii. If
→ a ⋅ 6 1
→
=
b
= 6 + 8x = 0,
8 x
,
then x =
iii. By computation,
4 3
.
then x = − 34 . Hence, when x = − 34 ,
Hence, when x =
π 4
,
3
,
→ a ∥
→ → a ⋅ b → → || a |||| b ||
=
→ b .
→
b .
− − − − − − → → → 2 2 a ⋅ b = 6 + 8x, || a || = √6 + 8 = 10,
cos θ =
If θ =
4
→ a ⊥
→
and ||
− − − − − 2 b || = √1 + x
Hence,
6 + 8x
3 + 4x = . − − − − − − − − − − 2 2 10√ 1 + x 5√ 1 + x
then 3 + 4x
π
= cos = − − − − − 2 4 5√ 1 + x
– √2
.
(A.1.2)
2
This implies − − − − − – 2 2(3 + 4x) = √2 (5√ 1 + x ) .
Squaring both sides of the last equation implies https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
6/10
Solution 4 (3 + 4x)
2
2
= 2(25)(1 + x ).
Hence, 2
2
2(9 + 24x + 16x ) = 25 + 25x .
This implies that 7x2
+ 48x − 7 = 0.
Solving the equation, we obtain x =
1 7
and x = −7.
Note that x = −7 is not a solution of the equation (A.1.2) because it does not satisfy 3 + 4x > 0.
12. 12.
Hence, when x =
1 7
,
the angle between
→ a
and
→ b
is
π 4
.
By computation, we have − − − →
2
||P 1 P 2 || − − − →
2
||P 2 P 3 || − − − →
2
||P 3 P 1 ||
2
= ||(0, 2, 1, − 3)||
2
= 0
2
= ||(−1, − 1, − 2, 1)|| 2
= ||(1, − 1, 1, 2)||
2
= 1
− − − →
2
||P 1 P 2 ||
2
+2
2
+1
= (−1)
2
+ (−1)
+ (−3)
+ (−1)
2
2
+1
− − − →
2
= ||P 2 P 3 ||
2
2
= 14.
+ (−2) 2
+2
2
2
+1
= 7 and
= 7,
− − − →
2
+ ||P 3 P 1 ||
By the Pythagorean theorem, the triangle is a right triangle and − − − →
− − − →
P2 P3 ⊥ P3 P1 .
− − − →
b. Because P2 P3
− − − → ⊥ P3 P1 ,
the area of the triangle is 1 2
− − − →
− − − →
||P 2 P 3 ||||P 3 P 1 || =
1
7 – – √7√7 = . 2 2
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Solution
a. By computation, we have − − → AB = (6, 3, 3, 9) − (4, 5, 5, 7) = (2,
− 2,
− 2, 2),
− − → CD = (3, 4, 4, 6) − (5, 2, 2, 8) = (−2, 2, 2,
− 2),
− − → AD = (3, 4, 4, 6) − (4, 5, 5, 7) = (1, 1, 1, 1), − − → BC = (5, 2, 2, 8) − (6, 3, 3, 9) = (−1,
− − →
− − →
− − →
− − →
− 1,
− − → − − →
− 1,
− − →
− 1).
− − →
It is easy to see that AB = −CD, AD = −BC . Hence AB||CD and AD||BC and the quadrilateral is a parallelogram. Because − − →
− − →
AD ⋅ CD = (1, 1, 1, 1) ⋅ (−2, 2, 2, − 2) = −2 + 2 + 2 − 2 = 0,
− − →
− − →
and the quadrilateral is a rectangle. b. The area of the rectangle is AD ⊥ CD
− − → − − → − − − − − − − − − − − − − − − − − − − − − − ||AD||||CD|| = √ 1 + 1 + 1 + 1 ⋅ √ 4 + 4 + 4 + 4 = 8.
Section 1.3
→
→ a
and
→
→
( b − proj→ a
→ b ) ⊥ a .
b
, find Proj→
N
1. For each pair of vectors
1.
→ → a = (−1, 2), b = (1, 1);
2.
→ a = (2,
3.
→ a = (−1,
→
a
b
and verify that
→ − 1, 1), b
= (1,
− 1, 2);
→ − 2, 2),
b
= (1, 0,
− 1);
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Solution
4. 2. Let
3. Let
→ a = (0,
→ − 1, 1,
→ a = (1, 1, 1)
and
− 1),
b
= (1, 1,
− 1, 2).
→ b
= (1, −1, 1). ∥
→∥
1. Use Theorem 1.3.2
(i) to find ∥Pr oj→
2. Use Theorem 1.3.2
(ii) to find cos θ, where θ is the angle between
→ a = (1, 0, 1, 2)
and
b ∥.
a
→ b
= (−
1 2
,
1 2
,
1 2
,−
3 2
∥
2. Use Theorem 1.3.2
(ii) to find θ, where θ is the angle between
→ a = (1,0),
2.
→ a = (1,0),
→ b .
→∥
(i) to find ∥Pr oj→
1.
and
).
1. Use Theorem 1.3.2
b ∥.
a
4. Find the area of the parallelogram determined by
→ a
→ a
and
→ a
and
→ a
and
→ b .
→ b .
→ b
= (0, −1);
→ b
= (−1, 1);
→
3.
→ a = (0, 2), b
4.
→ a = (1,
= (1, 1); →
− 1),
b
= (−1, 2).
5. Find the area of the parallelogram determined by
→ b .
→
1.
→ a = (1, 0, 1),
2.
→ → a = (1, 0, 0), b = (−1, 1, − 2);
3.
→ a = (0, 2,
4.
→ → a = (1, − 1, − 1), b = (−1, 1, 2);
5.
→ → a = (1, 0, 2, 1), b = (2, 1, 0, 3);
b
= (−1, 0, 1);
→ − 1), b
= (−5, 1, 1);
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Solution
6.
→ → a = (−2, − 1, 0, 3), b = (3, 1, − 1, 0).
→
6. Find G( a
→ , b )
→ → ∥ ∥ b − Porj→ b ∥ a
and ∥
for each pair of vectors.
a.
→ → a = (−1, 2), b = (1, − 1);
b.
→ → a = (2, − 1, 1), b = (1, − 1, 2);
c.
→ → a = (1, 2, 0, 1), b
d.
→ a
= (1, − 1, 2, 3); →
= (2, − 1, 1, 0),
b
= (3, 0, 1, 3).
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Solution
Solution 1. 1. 1.
→
→ Proj→ b
= (
a
Because
→ b ⋅ a 2
∣∣→∣∣ a ∣∣ ∣∣
→ ) a =
(1)(−1)+(1)(2) 2
2
(−1, 2) = (−
(−1) +2
→
→
b − Proj→ b a
= (1, 1) − (−
1 5
2
,
5
) = (
→ → 6 → ( b − Proj→ b ) ⋅ a = ( , a 5
6 5
3
,
3 5
1 5
,
2 5
).
),
) ⋅ (−1, 2) = −
5
6
6
+
5
= 0
5
→
and ( 2.
→ → b − Proj→ b ) ⊥ a . a
→
→ Proj→ b
= (
a
Because
→ b ⋅ a 2
∣∣→∣∣ a ∣∣ ∣∣
→ ) a =
(1)(2)+(−1)(−1)+(2)(1) 2
2
2
(2, − 1, 1) = (
2 +(−1) +1
→
→
b − Proj→ b a
= (1, − 1, 2) − (
5
, −
3
5 6
,
5 6
) = (−
2 3
5 3
, −
, −
1 6
5
,
6
,
7 6
5 6
).
),
→ → 2 1 7 4 1 7 → ( b − Proj→ b ) ⋅ a = (− , − , ) ⋅ (2, − 1, 1) = − + + = 0 a 3 6 6 3 6 6
→
and ( 3.
→ → b − Proj→ b ) ⊥ a . a
a
Because
…
1/7
→
→ Proj→ b
= (
→ b ⋅ a 2
∣∣→∣∣ a ∣∣ ∣∣
→ ) a =
(1)(−1)+(0)(−2)+(−1)(2) 2
2
2
(−1, − 2, 2) = (
(−1) +(−2) +2
→
→
b − Proj→ b a
= (1, 0, − 1) − (
1 3
,
2 3
, −
2 3
) = (
2 3
, −
2 3
1 3
,
2 3
, −
, −
1 3
),
2 3
).
Solution →
→
2 2 1 2 4 2 → ( b − Proj→ b ) ⋅ a = ( , − , − ) ⋅ (−1, − 2, 2) = − + − = 0 a 3 3 3 3 3 3
→
and (
→ → b − Proj→ b ) ⊥ a . a
→
→ Proj→ b
= (
a
4.
=
→ b ⋅ a
→ 2 ∥ a ∥
→ ) a
(1)(0)+(1)(−1)+(−1)(1)+2(−1) 2
2
2
2
(0, − 1, 1,
− 1)
0 +(−1) +1 +(−1)
=
Because
→
−4 3
(0, − 1, 1, − 1) = (0,
→
b − Proj→ b a
= (1, 1, − 1, 2) − (0,
→ → 1 1 → ( b − Proj→ b ) ⋅ a = (1, − , , a 3 3
2
4 3
, −
4 3
,
4 3
4 3
, −
4 3
,
) = (1, −
4 3
1 3
).
,
) ⋅ (0, − 1, 1, − 1) = 0 +
3
1 3
,
1 3
2 3
+
),
1
−
3
→
and (
→ → b − Proj→ b ) ⊥ a . a
2. 2. →
→
→
1. Because a ⋅ b = (1)(1) + (1)(−1) + (1)(1) = 1 and ∥ a by Theorem 1.3.2 (i), ∣→ →∣ ∣ a ⋅ b ∣ ∣ ∣
→ ∥Proj→ b a
→
2. Because ∥ …
2/7
b
− − − − − − − − − − − − − 2 ∥= √1 + (−1)
2
2
+1
∥=
→ ∥ a ∥
– = √3,
|1| =
− − − − − − − − − − – 2 2 2 ∥= √1 + 1 + 1 = √3,
1
– = – = √3 √3
by Theorem 1.3.2 (ii),
– √3 3
.
2 3
= 0
Solution →
√3
∥Proj→ b ∥ a
|cos θ| =
=
→ ∥ b ∥
Because
→ a ⋅
→ b
= 1 > 0, cos θ > 0
and cos
1 . – = 3 √3 3
∣ θ = ∣ ∣ cos θ∣ =
1 3
.
3. 3. 1. Because 1 1 1 3 → → a ⋅ b = (1) (− ) + (0) (− ) + (1) ( ) + (2) (− ) = −3 2 2 2 2
→
and || a
− − − − − − − − − − − − − − – 2 2 2 2 || = √1 + 0 + 1 + 2 = √6,
by Theorem 1.3.2 (i), ∣→ ∣ a ⋅ ∣
→ ||Proj→ b || =
∣ b ∣ ∣
|−3| =
→ ∣∣ ∣∣ a ∣∣ ∣∣
a
2. Because
→
– √6
− −−−− −−−−− −−−− −−−− −−−−− −
→
|| b || = √(−
1 2
2
)
+(
1 2
2
)
+(
1 2
2
)
+ (−
3 2
2
)
=
3 – √6
=
– √6
.
2
− − −− − −− − −− − − = √
1 4
+
1 4
+
1 4
+
9 4
– = √3.
Theorem 1.3.2 (ii), →
√6
||Proj→ b || |cos θ| =
a
→ || b ||
Because θ = π−
…
3/7
→ → a ⋅ b = −3 < 0, cos θ < 0. π 4
=
3π 4
.
Hence, cos
=
2
3
– √2 – √3 = . 2
∣ ∣ θ = − cos θ = − ∣ ∣
√2 2
.
Hence,
by
Solution
4. 4. 1.
∣∣1 → → A( a , b ) = ∣ ∣ ∣∣0
2.
∣∣ 1 → → A( a , b ) = ∣ ∣ ∣ ∣ −1
3.
∣∣0 → → A( a , b ) = ∣ ∣ ∣∣1
4.
∣∣ 1 → → A( a , b ) = ∣ ∣ ∣ ∣ −1
∣∣ ∣ ∣ = | − 1| = 1; −1 ∣ ∣ 0
0∣∣ ∣ ∣ = |1| = 1; 1∣∣ 2∣∣ ∣ ∣ = | − 2| = 2; 1∣∣ −1 ∣ ∣ ∣ ∣ = |1| = 1; 2 ∣∣
5. 5. 1. Because ∣ 1 → → G( a , b ) = ∣ ∣ −1 → A( a ,
2.
∣ 1 → → G( a , b ) = ∣ ∣ −1
→ b ) =
0∣ ∣ 1∣
2
0∣ ∣ 0∣
2
∣ 1 + ∣ ∣ −1
1∣ ∣ 1∣
2
∣0 + ∣ ∣0
1∣ ∣ 1∣
2 2
= 0
2
+2
2
+0
− − − − − − − − → → – G( a , b ) = √4 = 2.
√
∣ 1 +∣ ∣ −1
∣ ∣ −2 ∣ 0
2
∣0 +∣ ∣1
∣ ∣ −2 ∣
∣2 +∣ ∣1
−1 ∣ ∣ 1 ∣
0
2 2
= 1
+ (−2)
2
2
+0
= 5,
− − − − − − − − → → → → – √ A( a , b ) = G( a , b ) = √5.
3.
∣ 0 → → G( a , b ) = ∣ ∣ −5
2∣ ∣ 1∣
2
∣ 0 +∣ ∣ −5
−1 ∣ ∣ 1 ∣
− − − − − − − − → → → → − − − √ A( a , b ) = G( a , b ) = √134.
…
4/7
2
2
= 10
2
2
+5
2
+3
= 134,
= 4,
Solution → G( a ,
→
∣ b ) = ∣
4.
−1 ∣ ∣
1
∣ −1 → A( a ,
→ b ) =
1
2
∣
∣ +∣
−1 ∣ ∣
1
∣ −1
2
2
∣ −1 +∣
∣
−1 ∣ ∣
1
∣
2
2 2
= 0
2
+1
+ (−1)
2
= 2,
∣
− − − − − − − − → → – G( a , b ) = √2.
√
→ → → → 2 → 2 , b ) = || a || || b || − |( a ⋅ b ) = (6)(14) − 25 = 39, − − − − − − − − → → → → − − √ A( a , b ) = G( a , b ) = √39. →
5. Because G( a
by Theorem 1.3.3,
6. Because → G( a ,
by Theorem 1.3.3,
→
→ → → 2 → 2 2 b ) = || a || || b || − |( a ⋅ b ) = (14)(11) − (−7) = 105,
− − − − − − − − → → → → − − − √ A( a , b ) = G( a , b ) = √105.
6. 6. 1. By Theorem 1.3.3 and Corollary 1.3.1,
r
→ → 2 ∣ −1 → → G( a , b ) = [A( a , b )] = (∣
→ → √ G( a , b )
→∣∣
tf
∣∣→
∣∣ b − Proj→ b ∣∣ = ∣∣
a
→ ∣∣ ∣∣ a ∣∣ ∣∣
∣∣
∣
2
1
2
= 1.
−1 ∣
1
=
∣ ∣)
=
√5
√5 5
.
2. By Theorem 1.3.3 and Corollary 1.3.1, ∣2 → → G( a , b ) = ∣ ∣1 = (−1)
…
5/7
−1 ∣ ∣ −1 ∣ 2
2
2
+3
∣2 + ∣ ∣1
1∣ ∣ 2∣
+ (−1)
2
2
∣ −1 + ∣ ∣ −1
= 11
1∣ ∣ 2∣
2
Solution →
and ||
→
b − Proj→ b || =
→ → √ G( a , b ) → || a ||
a
√11
=
2
2
√66
=
6
2
√2 +(−1) +1
.
3. By Theorem 1.3.3, → G( a ,
→
→ → → 2 → 2 2 b ) = || a || || b || − |( a ⋅ b )| = (6)(15) − 4 = 86.
→
→
|| b − Proj→ b || =
→ → √ G( a , b ) → || a ||
a
√86
=
2
2
2
=
√86 2
√1 +2 +0+1
.
4. By Theorem 1.3.3, →
→ → → 2 → 2 2 b ) = || a || || b || − |( a ⋅ b )| = (6)(19) − 49 = 65.
→ G( a ,
→
→ → √ G( a , b )
→
|| b − Proj→ b || a
=
=
=
→ || a ||
√65 √6
=
2
√22 +(−1) +12 +0
√390 6
.
Section 1.4 1. Let
1 1 ⎛ ⎞ ⎛ ⎞ → → a1 = ⎜ 0 ⎟ , a2 = ⎜ 3 ⎟ , ⎝
i. Let ii. Let iii. Let
…
6/7
1
⎠
⎝
2
and
⎠
→ b
= (−1,
− 9,
→ b
= (3, 6, 6)
→ b
= (2, 7, 7)
T
T
− 4)
⎛ → a3 = ⎜ ⎝
T
.
−1 1 3
⎞ ⎟. ⎠ →
Verify whether 2a1
→ → − 3a2 = b .
→
→ → − 2a2 = b .
→
→ → − 2a2 + a3 =
.
Verify whether a1
.
Verify whether a1
√65
→ b .
Solution
2. Let
1 1 → → a1 = ( ) , a2 = ( ), 2 1
and
→ b
1 = (
).
Determine whether
→ b
is a linear combination of
0
→ − → a1 , a2 . →
3. Let a1
= (
1
)
→
and a2
= (
1
4. Let
2. Is
→ 0 = (0, 0, 1) → b
= (1, 0, 2)
7/7
Show whether
→ b
= (
1
0
)
→
1 →
and a4
→
→
→
→
→
→
→
→
a linear combination of a1 , a2 , a3 , a4 ?
T
a linear combination of a1 , a2 , a3 , a4 ? 2 −1
),
and
→ b
= (
4 3
).
→
is a linear combination of a1 , a2 .
T
1 → → a1 = ( ) , a2 = ( 2
…
).
→ → → T T T a1 = (1, − 1, 1) , a2 = (0, 1, 1) , a3 = (1, 3, 2) ,
1. Is
5. Let
2
Show that
→ b
= (0, 0, 1)
T
.
→ → ∈ span{a1 , a2 }.
Solution
Solution 1. 1. 1 1 2 3 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ → → 2a1 − 3a2 = 2 ⎜ 0 ⎟ − 3 ⎜ 3 ⎟ = ⎜ 0 ⎟ − ⎜ 9 ⎟ ⎝
i. ⎛
2−3
⎞
1
⎛
⎠
⎝
−1
⎞
ii.
2−6
⎠
⎝
−4
⎠
⎝
2
⎠
⎝
6
⎠
→
= ⎜ 0 − 9 ⎟ = ⎜ −9 ⎟ = ⎝
2
b .
⎠
1 1 1−2 −1 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ → → → a1 − 2a2 = ⎜ 0 ⎟ − 2 ⎜ 3 ⎟ = ⎜ 0 − 6 ⎟ = ⎜ −6 ⎟ ≠ b . ⎝
1
⎠
⎝
2
⎠
⎝
1−4
⎠
⎝
−3
⎠
1 1 −1 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ → → → a1 + 2a2 + a3 = ⎜ 0 ⎟ + 2 ⎜ 3 ⎟ + ⎜ 1 ⎟ ⎝
iii. ⎛
1+2−1
⎞
1
⎠
⎛
2
⎝
⎞
2. 2.
1+4+3
→
Let xa1
⎠
→ → + ya2 = b .
⎝
8
⎠
⎝
⎠
Then x (
1
1 ) + y(
2
1/3
y )+(
2x
1 ) = (
1
x
…
⎠
b .
(
Hence,
3
→
= ⎜0 + 6 + 1 ⎟ = ⎜7⎟ ≠ ⎝
2
)
and
0
x+y ) = (
y
1 ) = (
2x + y
). 0
Solution
(1) − (2) : (x + y) − (2x + y) = 1 − 0
x+y = 1
(1)
2x + y = 0.
(2)
and x = −1. By (1),
y = 1 − x = 1 − (−1) = 1 + 1 = 2.
→
Hence, −a1 3. 3.
→ + 2a2 =
→
Let xa1
→ b
and
→ → + ya2 = b .
→ b
→
→
is a linear combination of a1 and a2 .
Then x(
1
) + y(
1
2
) = (
1
0
).
1
Hence,
→
Hence, 2a1
Let
By Theorem 1.4.2,
→ b
→ → − a2 = b
→ → → x1 a1 + x2 a2 = b .
→ b
→ b
→
→
→
→
Then 1
) + x2 (
2 −1
→
is a linear combination of a1 and a2 .
→ → → = a1 + a2 , b
→
2
2/3
and
is a linear combination of a1 , a2 , a3 , a4 .
x1 (
…
(2)
The result (1) follows from Theorem 1.4.1. Because
→ − → a1 , a2 .
5. 5.
x + y = 1.
implies that (x + 2y) − (x + y) = 0 − 1 and y = −1. By
(1), x = −2y = −2(−1) = 2.
4. 4.
(1)
t
(1) − (2)
x + 2y = 0
) = (
4 3
).
is a linear combination of
Solution
This implies
implies x2
…
3/3
= 10
= 2x1 − 3 = 2(2) − 3 = 1.
(1)
2x1 − x2 = 3.
(2)
and x1
Hence,
= 2.
→ b
Substituting x1
→ → = 2a1 + a2
and
= 2
→ b
into the second equation
→ → ∈ span{a1 , a2 }.
n
Equation (1) + (2) × 2 implies 5x1
x1 + 2x2 = 4
A.2 Matrices
A.2 Matrices Section 2.1 1. Find the sizes of the following matrices: ⎛ A = (2) B = (
5
8
1
4
)
C = (
10
3
10
3
1
0
8
⎜0 ) D = ⎜ ⎜0 4 ⎝
0
10
2 ⎟ ⎟ 1 ⎟
0
6
⎞
9
10
⎠
2. Use column vectors and row vectors to rewrite each of the following matrices. 3
2
A = ⎜1
8
1 ⎟
⎛
⎝
5
4
10
⎞
⎛
1
B = ⎜3
⎠
⎝
⎛
9
7
2
10
8
7⎟
3
10
4
4
⎞
9
8
⎜6 C = ⎜ ⎜3
⎠
⎝
7
⎞
is t
3
0
5
4
2
1
−1
−2
⎟ ⎟ ⎟ ⎠
3. Find the transposes of the following matrices: ⎛ C = (4) B = ( 3 11 2 )
9 6
5. Let C
= (
⎛
6. Let E
E = F.
1/3
2
)
and B = (
x
12
5
25
19
4
6
120
= ⎜ 123 ⎝
…
3
126
25
)
12
9
3
6
4
and D = (
122
⎞
C = ⎜ 8 ⎟ ⎝
4. Let A = (
33
124
125 ⎟
127
128
⎠
).
and F
3
9
D = ⎜ −2
⎠
⎝
5
−19
⎞
8
−7 ⎟
3
−9
⎠
Find all x ∈ R such that A = B. 2
12
5
x
19
4
6
⎛
⎞
⎛
= ⎜ ⎝
).
120
Find all x ∈ R such that C 2
x
123
124
26x − 4
127
122 3
x
128
= D.
⎞ ⎟. ⎠
Find all x ∈ R such that
A.2 Matrices a
b
3x + 2y
−x + y
7. Let A = (
)
and B = (
1
1
3
6
).
Find all a, b, x, y ∈ R such that A = B.
8. Let a+b
2b − a
x − 2y
5x + 3y
C = (
Find all a, b, x, y ∈ R such that C 9. Let A = (
3
4
6
−1
−8
)
= D.
and B = (
7
−8
9
0
−1
0
Compute
−A; 4A − 2B; 100A + B.
9
5
1
2
2
2
10. Let A = ⎜ 8
0
0 ⎟ , B = ⎜ 0
0
0⎟,
3
2
6
8
0
⎞
⎛
⎠
⎝
4
and C
⎠
Compute 1. 3A − 2B + C, 2.
[3(A + B)]
3.
(4A +
1 2
T
7
0
= ⎜0
3
1⎟.
0
2
⎞
⎝
⎠
0
,
B − C)
T
.
11. Find the matrix A if [(3AT ) − (
−7
−2 9
T
T
) −6
12. Find the matrix B if
4
⎛
⎞
or
⎝
2/3
).
A + B;
⎛
…
and D = (
D
i. ii. iii. iv.
−2
)
]
= (
−5
−10
33
12
).
6
0
8
14
).
A.2 Matrices ⎡
1
⎢ ⎣
…
3/3
2
⎛
6
B + ⎜8 ⎝
1
3
T
⎞⎤
⎛
3 ⎟⎥
− 3⎜
4
⎠⎦
⎝
−5 8 −4
6
T
⎞
−9 ⎟ 2
⎠
= (
23
−16
17
−16
26.5
2
).
Solution
Solution 1. 1. 2. 2.
The sizes of A, B, C, and D are 1 × 1, 2 × 2, 2 × 3, and 4 × 3, respectively. Let
3 3 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ → → → c1 = ⎜ 1 ⎟ , c2 = ⎜ 8 ⎟ , c3 = ⎜ 1 ⎟ . ⎝
5
⎠
⎝
4
⎠
⎝
10
→
→ c2
Then A = ( c1
→ c3 ).
⎠
Let
→ r1 = (3
3
→ 2), r2 = (1
8
→ 1), r3 = (5
4
10).
ut
→ ⎛ r1 ⎞ →⎟ . Then A = ⎜ ⎜ r2 ⎟ ⎝ →⎠ r3
Let
1 9 7 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ → → → → c1 = ⎜ 3 ⎟ , c2 = ⎜ 10 ⎟ , c3 = ⎜ 8 ⎟ , c4 = ⎜ 7 ⎟ . ⎝
4
⎠
⎝
3
⎠
⎝
10
⎠
⎝
4
→
Then B = ( c1
→ c2
→ c3
→ c4 ).
⎠ → ⎛ r1 ⎞
Let
→ r1 = (1
9
7
→ 2), r2 = (3
10
8
→ 7), r3 = (4
3
10
4).
→⎟ . Then B = ⎜ ⎜ r2 ⎟ ⎝ →⎠ r3
Let
9
⎞
⎛
→ ⎜6⎟ → ⎜ c1 = ⎜ ⎟ , c2 = ⎜ ⎜3⎟ ⎜ ⎝
0
⎠
⎝
8 5 2 −1
⎞ ⎟ ⎟, ⎟ ⎠
⎛
and
7
tf
⎛
→ ⎜ c3 = ⎜ ⎜ ⎝
4 1
−2
⎞
⎟ ⎟. ⎟
Then C
→ = ( c1
→ c2
→ c3 ).
⎠
https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
1/10
Solution
Let
→ r1 = (9
→ 7), r2 = (6
8
→ 4), r3 = (3
5
2
1),
and
→ r4 = (0
−1
− 2).
Then
→ ⎛ r1 ⎞ ⎜ ⎜ C = ⎜ ⎜ ⎜
→⎟ r2 ⎟ ⎟. →⎟ r3 ⎟
⎝ →⎠ r4 ⎛
3. 3.
T
A
T
= (4), B
3
⎞
= ⎜ 11 ⎟ , C ⎝
2
T
= (33
8
12),
and
⎠
⎛ D
T
= ⎜ ⎝
3
−2
5
9
8
3
−19
−7
−9
⎞ ⎟ ⎠
Let A = B. Then x2 = 4. This implies x = 2 or x = −2. Let C = D. Then x2 = 25. Solving the equation, we obtain x = 5 or x = −5. Let E = F . Then x2 = 25, x3 = 125, and 26x − 4 = 126. Solving the three equations implies 130 x = 5 or x = −5; x = 5; and x = = 5. Hence, when x = 5, E = F . 26
4. 4. 5. 5. 6. 6.
7. 7. (1)
Let A = B. Then a = 1 and b = 1. Hence,
3x + 2y = 3
(2) −x + y = 6.
(1) − (2) × 2
implies that 3x + 2y − 2(−x + y) = 3 − 12 and x = − 95 . By (2), we obtain
y = x+6 = −
9 5
+6 =
21 5
.
Hence, when a = 1, b = 1, x = − 95 , y =
21 5
, A = B.
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Solution
8. 8. (1)
Let C
= D.
Then
a+b = 6
(2) 2b − a = 0
(3) x − 2y = 8
(4) 5x + 3y = 14.
(1) + (2)
implies that (a + b) + (2b − a) = 6 + 0 = 6 and b = 2. By (1), we have
a = 6 − b = 6 − 2 = 4.
implies that 5(x − 2y) − (5x + 3y) = 40 − 14 and −13y = 26. Hence, y = −2. By (3), we get x = 8 + 2y = 8 + 2(−2) = 4. Hence, whena = 4, b = 2, x = 4, and (3) × 5 − (4)
y = −2, C = D.
9. 9. −2
3
4
A+B = ( 6
i.
7
−8
9
0
−1
0
)+( −1
−8
−2 + 7
3−8
)
4+9
= ( 6+0
ii.
−A = − (
−2
3
4
6
−1
−8
5
−5
13
6
−2
−8
) = (
) = (
−1 − 1
2
−3
−4
−6
1
8
−8 + 0
).
).
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Solution
iii.
−2
3
−8
12
−1
−8
16
14
= (
−16
9
0
−1
0
)
18
−4
−32
0 −2
(
28
−2
3
−2
0
24
−2
−32
4
).
7
−8
9
0
−1
0
)+( 6
(
−22 ) = (
100A + B = 100 (
=
−8
)−( 24
=
7 ) − 2(
6
iv.
4
4A − 2B = 4 (
−1
−200
300
400
600
−100
−800
−8
)+(
)
7
−8
9
0
−1
0
−200 + 7
300 − 8
400 + 9
600 + 0
−100 − 1
−800 + 0
)
) = (
−193
292
409
600
−101
−800
).
10. 10. 1. Let D = 3A − 2B + C. Then
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Solution 9
5
1
2
2
2
4
7
0
D = 3⎜8
0
0 ⎟ − 2⎜0
0
0 ⎟ + ⎜0
3
1⎟
3
2
6
8
0
2
⎛
⎝
⎛
0
27
15
= ⎜ 24 ⎝
⎛
9
23
⎛
⎝
−8
2.
[3(A + B)]
T
= 3(A + B)
T
0⎟ − ⎜0
0
0 ⎟ + ⎜0
3
1⎟
0
2
⎠
0
−3
−10
T
3−4
16
⎝
7
0
0 − 0 ⎟ + ⎜0
3
1⎟
0
2
6 − 16
⎠
⎛
⎝
0
4
7
0
⎟ + ⎜0
3
1⎟
0
2
⎞
⎛
⎠
⎝
0
−1 + 0
0+3
0+1
−3 + 0
−10 + 2
T
0
4
⎞
11 + 7
+B
⎠
⎞
⎛
27
⎝
−8
0
4
= 3 ⎢⎜ 5
0
3 ⎟ + ⎜2
0
6 ⎟⎥ = 3 ⎜ 7
33
= ⎜ 21 ⎝
B − C)
T
9
.
1
0
24
2 12
⎠
⎞
⎠
18
−1
3
1
−3
−8
⎞ ⎟. ⎠
)
2
⎞
⎠
⎠
⎟ = ⎜ 24 ⎠
⎞
⎞
0
⎛
2
12
⎛
8
⎣⎝
1
8
⎞
9
⎡⎛
3. Let D = (4A +
⎝
0
= 3(A
⎠
0
−1
−8 + 0
0
7
= ⎜ 24 + 0 ⎝
⎝
4
9 − 12
23 + 4
⎛
⎠
⎞
4
11
= ⎜ 24
⎛
4
0−0
0−8
4
⎞
4
⎛
15 − 4
= ⎜ 24 − 0 ⎝
⎝
⎞
6
27 − 4
⎛
⎠
3
0
0
⎞
⎛
⎝
2
0
8
⎞⎤
⎠⎦
⎛
⎝
11
3
8
4
0
9 ⎟
0
10
⎞
⎠
⎞
0
27 ⎟ .
0
30
⎠
Then
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Solution ⎛
9
D = 4⎜5 ⎝
⎛
1
36
= ⎜ 20 ⎝
⎛
4
8
0
2
0
4
4
0
0
⎜2
0
6 ⎟ − ⎜7
3
0⎟
2
0
8
0
1
2
1
0
2
4
0
0
12 ⎟ + ⎜ 1
0
3 ⎟ − ⎜7
3
0⎟
0
1
2
4
0
0
12 + 3 ⎟ − ⎜ 7
3
0⎟
1
2
⎞
0
3⎟+
0
2
32
0
0 0
⎛
= ⎜ 21 ⎝
⎛
5
T
[(3A
)−(
−7
) −6
11. 11. T
= 3(A
)
T
−(
⎠
⎝
2
−2
−6
9
⎝
⎞
⎛
⎠
4
⎠
⎝
0
0
0
15 ⎟ − ⎜ 7
3
0⎟
0
12
1
2
37 − 4
⎛
⎠
⎝
32 − 0
0
]
T
= (3A
⎠
⎞
⎠
⎞
⎠
⎞
⎠
32
0−3
15 − 0 ⎟ = ⎜ 14
−3
15 ⎟ .
0−1
12 − 2
−1
10
T
−7
−2
−6
9
)
) = 3A − (
0
⎞
33
5−0
2−0
⎝
⎛
4
⎞
⎛
⎠
⎞
8+4
− [(
9
−7
0
0+2
0+0
32
1
⎞
⎞
−7
−2
−6
9
⎝
5
2
⎞
⎠
T
T
)
) = (
⎛
⎠
T
T
−2
⎛
0+0
= ⎜ 21 − 7 ⎝
⎝
32 + 0
4+1 37
2
⎞
8
36 + 1
1
⎠
= ⎜ 20 + 1 ⎝
⎛
]
−5
−10
33
12
).
Hence, 3A = (
−5
−10
33
12
)+(
−7
−2
−6
9
) = (
−12
−12
27
21
).
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Solution 1 A =
−12
⎢
⎛ 1
12. 12.
=
3
B + ⎜8
2
⎣
=
6
⎝
1
T
B
2
1
⎛
3 ⎟⎥
− 3⎜
⎠⎦
⎝
4 6
8
T
B
2
27
3
8
−4
6
−9
2
)
1 + 12
1
) = 3 + 27
17
−16
26.5
2
).
⎠
2
−5
8 − 24
−16
7
⎞
+(
23
9
−9 ⎟
−4
4
6 + 15
−4
T
6
8
21
) − 3(
3 − 18
= (
−5
1
+( 3
1
T
⎞⎤
−4 ) = (
3
⎡
−12
(
2
4−6
T
B
21
−16
13
−15
30
−2
+(
)
).
This implies that 1 2
T
B
23
−16
B
21
−16
13
)−( −16
T
17
= ( 26.5
2
2
0
4
−1
−3.5
4
= 2(
−15
) = (
2
0
4
−1
−3.5
4
) = ( 30
4
0
8
−2
−7
8
−2
),
⎛ )
4
and B = ⎜ 0 ⎝
8
−2
⎞
−7 ⎟ . 8
⎠
Section 2.2 1. Let A = (
2 1
and
1 −1
1 2
→
⎛
x1
⎞
− →
⎛
0
⎞
− →
⎛
2
⎞
) , X = ⎜ x2 ⎟ , X1 = ⎜ 1 ⎟ , X2 = ⎜ 1 ⎟ . ⎝
x3
⎠
⎝
2
⎠
⎝
1
→
− →
Compute A X , AX1 ,
⎠
− → AX 2 .
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Solution ⎛
−2
−1
3
1
0
1
2. 2. A = ⎜ ⎝
⎛
0
3. Let A = ⎜ 1 ⎝
1
⎞
→
⎛
− →
1
⎠
⎝
1
→
⎞
− →
2
−1
1
⎞
− → ⎜1⎟ ) , X3 = ⎜ ⎟. ⎜1⎟ 0 1
⎟ , X2 = (
0
− →
Compute A X and AX1 .
).
⎛
1
−1 ⎟ , X1 = ⎜
2
) , X1 = (
2a
⎠
⎞
− →
−a
⎟, X = (
⎠
⎝
2
⎠
− →
Determine whether AXi . is defined for each i = 1, 2, 3. 1
4. Let A = ⎜ 1
0
⎝
0
⎞
−1 ⎟ , X = ⎜ ⎠
1
1
2
1
1
−1
5. Let A = ⎜ 1
0
2
1
−1
⎛
⎝
2
1
⎛
→
⎝
⎞ ⎟
and
0 −1 ⎛
→
⎞
→
⎟,
Y
= ⎜
⎠ 1
⎝
⎝
0
0 −1
⎞
⎞ ⎟.
→
Compute A( X
→
− 3 Y ).
⎠
→
X = ⎜ −1 ⎟ .
⎠
3
⎛
u
0
⎛
Write A X as a linear combination of the column
⎠
vectors of A. 6. We are interested in predicting the population change among three cities in a country. It is known that in the current year, 20% and 30% of the populations of City 1 will move to Cities 2 and 3, respectively; 10% and 20% of the populations of City 2 will move to Cities 1 and 3, respectively; and 25% and 10% of the populations of City 3 will move to Cities 1 and 2, respectively. Assume that 200,000 600,000, and 500,000 thousand people live in Cities, 1, 2, and 3, respectively, in the current year. Find the populations in the three cities in the following year. 7. Let A, B, and C be matrices such that AB = C. Assume that the sizes of B and C are 6 × 8 and 3 × 8, respectively. Find the size of A. 8. Let A = (
9. Let A = (
2
1
−2
3
)
and B = (
0
1
−1
2
3
1
1
−2
1
3
4
1
2
1
−1
1
0
1
−1
2
4
⎛ )
and B = ⎜ ⎝
).
Find AB and the sizes of A, B, and AB. ⎞ ⎟.
Compute AB and find the sizes of A, B, and AB.
⎠
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Solution
10. Let A = (
1
0
−2
3
2
−1
0
⎛ )
−1
2
2
1
0
0
and B = ⎜ −1 ⎝
2
1
⎞
Compute AB and find the sizes of A,
−2 ⎟ . 1
⎠
B, and AB. ⎛
1
11. 10. Let A = ⎜ 1 ⎝
3
1
⎞
3⎟ 1
and B = (
⎠
2
1
−2
3
).
Are AB and BA defined?
).
Are AB and BA defined?
If so, compute them. If not, explain why. 12. Let A = (
2
−1 )
−2
and B = (
3
0
−2
4
2
If so, compute them. If not, explain why. Is AB equal to BA? 13. Let A = (
1
−1
−1
2
) , B = (
0
−1
2
1
1
3
1
−1
0
3
2⎟.
−2
0
3
⎛ ),
and C
= ⎜ ⎝
1
⎞
⎠
T
Show that A(BC) = (AB)C and (AB) = BT AT . 14. Two students buy three items in a bookstore. Students 1 and 2 buy 3,2,4 and 6,2,3, respectively. The unit prices and unit taxes are 5, 4, 6 (dollars) and 0.08, 0.07, 0.05, respectively. Use a matrix to show the total prices and taxes paid by each student. 15. Assume that three individuals have contracted a contagious disease and have had direct contacts with four people in a second group. The direct contacts can be expressed by a matrix 0
1
1
1
A = ⎜1
1
1
1⎟.
1
0
1
⎛
⎝
1
⎞
⎠
Now, assume that the second group has had a variety of direct contacts with five people in a third group. The direct contacts between groups 2 and 3 are expressed by a matrix
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Solution 1
0
0
1
0
⎜0 B = ⎜ ⎜1
0
0
1
1
0
1
0⎟ ⎟. 1⎟
0
0
1
0
⎛
⎝
1
⎞
⎠
i. Find the total number of indirect contacts between groups 1 and 3. ii. How many indirect contacts are between the second individual in group 1 and the fourth individual in group 3? 16. There are three product items, P1, P2, and P3, for sale from a large company. In the first day, 5, 10, and 100 are sought out. The corresponding unit profits are 500, 400, and 20 (in hundreds of dollars) and the corresponding unit taxes are 3, 2, and 1. Find the total profits and taxes in the first day.
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Solution
Solution AX
2
1
= ( 1
1. 1.
− → AX1 = (
− → AX2 = (
2. 2.
→
⎛
AX = ⎜ ⎝
−1
2
2
1
1
1
−1
2
2
1
1
1
−1
2
−2
−1
3
1
0
1
⎛
1
x1
⎞
) ⎜ x2 ⎟ = ( ⎝
⎛
x3 0
⎠
⎞
2x1 + x2 + x3
2(0) + 1(1) + 1(2)
)⎜1⎟ = ( ⎝
⎛
2 2
) = ( 1(0) + (−1)(1) + 2(2)
⎠
⎞
⎝
⎞ ⎟( ⎠
1
−a
) = ( 1(2) − 1(1) + 2(1)
⎠ ⎛
2a − 2a
⎞
2a
⎝
⎠
0 + 2a
⎛
−2
−1
3
1
4. 4.
→
1/8
6
− →
→
X − 3Y
⎛ = ⎜
1 0 −1
⎞
0
⎞
⎟(
1
⎠
1
2a
⎠
⎛
−4
) = ⎜
2
⎝
5 2
− →
⎛
⎟ − 3⎜ ⎠
).
3
is not defined, AX2 is defined, and AX3 is not defined.
⎝
…
0
⎝
⎞
tf ⎝
− →
).
) = ⎜ −3a + 2a ⎟ = ⎜ −a ⎟ .
AX 1 = ⎜
AX1
⎛
3 3
2(2) + 1(1) + 1(1)
)⎜1⎟ = (
− →
3. 3.
).
x1 − x2 + 2x3
ri
→
⎝
3 0 −1
⎞
⎛
⎟ = ⎜ ⎠
⎝
1−9 0−0 −1 + 3
⎞
⎛
⎟ = ⎜ ⎠
⎝
−8 0 2
⎞ ⎟. ⎠
⎞ ⎟. ⎠
Solution 0
1
A( X − 3 Y ) = ⎜ 1
0
→
⎛
→
⎝
5. 5.
1
1
−1
AX = ⎜ 1
0
2
1
−1
→
⎛
⎝
2
⎞⎛
1
⎞
⎛
1
1
0
−1 ⎟ ⎜
1
⎞
⎞⎛
⎠⎝
2
1
⎛
⎞
−8 0 2
⎛
⎟ ⎜ −1 ⎟ = ⎜ 1 ⎟ − ⎜ 0 ⎟ + 0 ⎜ ⎠⎝
0
⎠
⎝
2
⎠
⎝
1
⎠
⎝
⎞
⎛
0
⎞
⎟ = ⎜ −10 ⎟ . ⎠
−1
⎝
−4
⎠
⎞ ⎟.
2 −1
⎠
We denote by aij the percentage of populations moving from City j to City i, where i ≠ j and i, j = 1, 2, 3, and by aii the percentage of populations staying in City i. Let xi and yi denote the populations in Cities i in the current and following year, respectively. Then
6. 6.
y1
a11
a12
a13
50%
10%
25%
⎜ y2 ⎟ = ⎜ a21
a22
a23 ⎟ ⎜ x2 ⎟ = ⎜ 20%
70%
10% ⎟ ⎜ 600 ⎟
a32
a33
20%
65%
⎛
⎝
y3
⎛
⎞
⎝
⎠
⎛
a31
⎞⎛
⎠⎝
100 + 60 + 125
⎞
x1
x3
⎞
⎠
⎛
⎛
⎝
285
30%
⎞⎛
⎠⎝
200
500
⎞
⎠
⎞
= ⎜ 40 + 420 + 50 ⎟ = ⎜ 510 ⎟ . ⎝
60 + 120 + 325
⎠
⎝
505
⎠
Because AB = C and the sizes of B and C are 6 × 8 and 3 × 8, respectively, it follows from the 7. 7. formula (m × n)(n × r) = m × r that the size of A is 3 × 6. AB = (
8. 8.
2
1
−2
3
)(
2(1) + 1(3)
1
−2
1
3
4
1
)
2(−2) + 1(4)
2(1) + (1)
= (
) = ( −2(1) + 3(3)
−2(−2) + 3(4)
−2(1) + 3(1)
The sizes of A, B, AB are 2 × 2, 2 × 3, 2 × 3, respectively.
…
2/8
5
0
3
7
16
1
).
Solution
AB = (
9. 9.
0
1
−1
2
3
1
2
1
−1
1
0
1
−1
2
4
⎛ )⎜ ⎝
⎞ ⎟ ⎠
0(2) + 1(1) − 1(−1)
0(1) + 1(0) − 1(2)
0(−1) + 1(1) − 1(4)
2(2) + 3(1) + 1(−1)
2(1) + 3(0) + 1(2)
2(−1) + 3(1) + 1(4)
= (
= (
)
2
−2
−3
6
4
5
).
The sizes of A, B, AB are 2 × 3, 3 × 3, 2 × 3, respectively. 10. 10.
AB = (
1
0
3
2
−2
⎛
0
) ⎜ −1
−1
⎝
2
−1
2
2
1
0
0
1
⎞
−2 ⎟ = ( 1
⎠
−4
−1
2
−1
−4
1
8
−2
The sizes of A, B, AB are 2 × 3, 3 × 4, 2 × 4, respectively. ⎛
1
AB = ⎜ 1 ⎝
11. 11. ⎛
3 0
= ⎜ −4 ⎝
4
1
⎞
2
1
−2
3
3⎟( 1
⎠
4
⎛
1(2) + 1(−2)
) = ⎜ 1(2) + 3(−2) ⎝
3(2) + 1(−2)
1(1) + 1(3)
⎞
1(1) + 3(3) ⎟ 3(1) + 1(3)
⎠
⎞
10 ⎟ . 6
⎠
BA is not defined because the size of B is 2 × 2 and the size of A is 3 × 2.
…
3/8
).
Solution 2
−1
AB = ( −2
3
−4
−6
12
10
= (
12. 12.
0
0
4
−2
2
2
−1
)( 4
2
4
−6
4
2
= (
2(−2) − 1(2)
−2(0) + 3(4)
−2(−2) + 3(2)
0(2) − 2(−2)
0(−1) − 2(3)
4(2) + 2(−2)
4(−1) + 2(3)
)
) = ( −2
3
)
).
BC = (
0 1
A(BC) = (
13. 13. AB = (
(AB)C = (
−1 1
2 3
1
−1
−1
2
1
−1
−1
2
−1
−2
2
3
Hence, A(BC) = (AB)C.
4/8
2(0) − 1(4) ) = (
).
BA = (
…
−2
)(
⎛ )⎜ ⎝
)(
)(
−1
0
3
−2
0
1
3
−3
4
−5
2
12
0
−1
2
1
1
3
⎛ )⎜ ⎝
⎞
⎠
) = (
) = (
1
−1
0
3
−2
−4
−3
4
−5
2
12
1
−5
−8
−6
7
20
2⎟ = (
−4
−1 4
1
0
1
−1
−2
−1
2
3
4
⎞
2⎟ = ( 3
⎠
).
).
).
1
−5
−8
−6
7
20
).
Solution
T
B
T
A
= ⎜ −1 ⎝
2
1
⎞
3
⎠
0−1
⎛
T
(AB)
T
Hence, (AB) 14. 14.
T
= B
T
A
(
= (
−1
−1
2
0+2
= ⎜ −1 − 1 ⎝
1
1⎟(
2−3
⎞
−2 + 6
−2
−1
2
3
4
⎛
−1
2
⎟ = ⎜ −2
1+2
−1
)
⎠
T
)
⎝
⎛
= ⎜ −2 ⎝
3⎟.
−1
−1
−1
2
⎞
4
⎠
⎞
3⎟. 4
⎠
ut
0
⎛
.
Student 1
) = (
Student 2
3
2
4
6
2
3
3(5) + 2(4) + 4(6)
⎛
5
0.08
)⎜4 ⎝
⎞
0.07 ⎟
6
0.05
⎠
0.24 + 0.14 + 0.2
= (
) = ( 6(4) + 2(4) + 3(6)
0.48 + 0.14 + 0.15
47
0.58
56
0.77
).
15. 15. i. The total number of indirect contacts between groups 1 and 3 is given by
⎛
0
AB = ⎜ 1 ⎝
…
5/8
1
1 1 1
1 1 0
1
0
0
1
⎜0 1⎟⎜ ⎜1 ⎠ 1 ⎝ 1
0
0
1
1
0
1
1
⎛ ⎞
0
0
1
0
⎞
2 ⎛ 0⎟ ⎟ = ⎜3 1⎟ ⎝ 2 ⎠ 0
1
0
3
1
1
0
4
1⎟.
0
0
3
0
⎞
⎠
Solution
ii. From the last matrix, we see that c21 = 4, so there are four indirect contacts between the second individual in group 1 and the fourth in group 3.
⎛
500
16. 16. Let A = (5, 10, 100) and B = ⎜ 400 ⎝
20
3
⎞
2⎟. 1
Then the total profits and taxes in the first day can
⎠
be computed by 500
AB = (5, 10, 100) ⎜ 400 ⎝
20
3
⎞
2 ⎟ = (8500, 135) . 1
⎠
ut
⎛
Section 2.3 1. For each of the following matrices, find its trace and determine whether it is symmetric.
A1 = (
0 −1 −1
0
⎛
1
) , A2 = ⎜ 4 ⎝
5
4 −3 0
5
⎞
⎛
y 3
−1 ⎟ , A3 = ⎜ x 7
⎠
⎝
1
3
x
1
⎞
y
x⎟.
x
z
⎠
2. Identify which of the following matrices are upper triangular, diagonal, triangular, or an identity matrix.
…
6/8
Solution
A1 = (
0
2
0
0
⎛
1
A4 = ⎜ 1 ⎝
A7 = (
0 ).
−1 ⎛
2
4. Let A = ⎜ 0 ⎝
0
2
0
) A2 = ⎜ 2
0
0 ⎟ A3 = ⎜ 0
0
0⎟
0
1
0
0
0
1
0
1
) A6 = ⎜ 0
0
0⎟
0
1
⎝
x
0
⎞
0
1
1
0
⎛
⎝
1
0
0
) A8 = ⎜ 0
0
0 ⎟ A9 = ⎜
0
1
⎛
⎝
2
1
A10 = ⎜ 2
0
0 ⎟ A11 = ⎜ 0
0
1
1
⎝
⎠
0 ⎟ A5 = ( ⎠
⎛
⎞
0
0
⎛
⎞
⎝
⎠
0
0
⎝
⎠
⎠
⎞
⎠
1
0
0
0
2
0⎟
−1
1
1
⎛
⎞
0
0
⎞
−2 0
0
1
0 ⎟ A12 = ⎜ 0
1
0⎟
0
0
0
⎝
⎠
1
⎞
0⎟. 3
Find A6 .
⎠
− x − 4.
−1
1
1
2
).
Compute P(A).
Compute P(A) and P(B), where 3
⎝
4
1
0
−2
3
5
−1
⎞
⎛
⎟ and B = ⎜ ⎠
⎠
0
1 0
⎞
0
⎛
⎞
Find A0 , A2 , A3 , and A4 .
A = ⎜3
7/8
0
1
⎛
…
0
0
0
5. Let P (x) = 1 − 2x − x2 and A = ( 6. Let P (x) = x2
0
0
0
⎝
1
0
0
2
⎛
3. Let A = (
x
1
⎛
⎝
6
−2
−1
3
0
2
−1
3
4
⎞ ⎟. ⎠
⎞
⎠
Solution
7. Assume that the population of a species has a life span of less than 3 years. The females are divided into three age groups: group 1 contains those of ages less than 1, group 2 contains those of ages between 1 to 2, and group 3 contains those of age 2. Suppose that the females in group 1 do not give birth, and the average numbers of newborn females produced by one female in group 2 and 3, respectively, are 3 and 2. Suppose that 50% of the females in group 1 survive to age 1 and 80% of the females in group 2 live to age 2. Assume that the numbers of females in groups 1, 2, and 3 are X1 , X2 , and X3 , respectively. i. Find the number of females in the three groups in the following year. ii. Find the number of females in the three groups in n years. iii. If the current population distribution (x1 , x2 , x3 ) = (1000,500,100), then use result 2 to predict the population distribution in 2 years.
…
8/8
Solution
Solution 1. 1. i.
tr(A1 ) = 0 + 0 = 0, tr(A2 ) = 1 + (−3) + 7 = 5
and
tr(A3 ) = y + y + z = 2y + z.
ii.
A1
is symmetric, A2 is not symmetric, and A3 is symmetric.
Lower triangular: A2 , A3 , A7 , A8 , A9 . 2. 2. Upper triangular: A1 , A3 , A6 , A7 . Diagonal: A3 , A7 . Triangular: A1 , A2 , A3 , A6 , A7 , A8 , A9 . There are no identity matrices. 0
A
2
A
3
4
A
0
−1
1
1
0
−1
1
= (
= (
3. 3. A
1
2
= A
3
= A
0
)
= (
4. 4.
6
A
= ⎜0 ⎝
…
1/7
2
0
0
1
0
−1
1
1
0
−2
1
1
0
⋅A = (
0 −2 0
0
) = (
1
0
−2
1
1
0
−1
1
1
0
)(
1
⎠
6
⎛
2
= ⎜ 0 ⎝
0
).
) = (
1
0
−3
1
1
0
−4
1
) = ( −1
6
⎞
0⎟ 3
)
)( −3
⎛
0
1
)(
⋅A = (
1
1 0
(−2) 0
0 6
).
).
⎞
⎛
64
0 ⎟ = ⎜ 0 6
3
⎠
⎝
0
0
0
64
0
0
729
⎞ ⎟. ⎠
Solution
2
P (A) = I − 2A − A
5. 5.
= (
1
0
0
1
3
) − 2(
2
P (A) = A
⎛
1 1
1
2 2
⎝
⎛
3
4 12
= ⎜ 15 ⎝
⎛
)−(
1
1
2
−1
1
1
2
1 5
)−(
)(
1
−3
−3
−8
) = ( 1
1
0
−2
3
5 1 22
−11
5
0
19
2
⎞
⎛
⎠
3
3
− ⎜3
⎟
−1
23
= ⎜ 12 ⎝
−3
) − 2(
−1
−1
1
1
2
−1
1
1
2
2
)
)
).
− A − 4I
= ⎜3
2/7
0
)−( −2
…
0
−1
−2
= (
6. 6.
= (
1
⎝
⎞
4
⎛
⎠
20
−12
−16
13
⎝
0
−2
3
5 3
−9 ⎟ − ⎜ 3 16
1
4
1
0
0
⎟ − 4⎜0
1
0⎟
0
0
1
4
0
0
⎟ − ⎜0
4
0⎟
0
4
⎞
−1
⎠
1
0
−2
3
5
−1
⎞
⎠
⎛
⎝
⎛
⎝
0
⎞
⎠
⎞
⎠
Solution 2
P (B) = B
6
−2
−1
3
0
2
−1
3
4
⎛ = ⎜ ⎝
− B − 4I
31
⎛
⎛
= ⎜ 13 ⎝
0
⎟
−⎜
⎠
⎝
−14
0
5
14
23
−13
−13
−4
3
11
15
−1 21
⎛
−15
= ⎜ 16 ⎝
2
⎞
⎞
6
−2
−1
3
0
2
−1
3
4
6
−2
−1
3
0
2
−1
3
4
⎛
⎟−⎜ ⎠
⎝
1
0
0
⎟ − 4⎜0
1
0⎟
0
0
1
4
0
0
⎟ − ⎜0
4
0⎟
0
4
⎞
⎛
⎠
⎝
⎞
⎠
⎛
⎝
0
⎞
⎠
⎞
⎠
⎞ ⎟. ⎠
7. 7. i. We denote by y1 , y2 , and y3 the numbers of females in groups 1, 2, and 3, respectively. Then the first group in the following year consists of the newborn females from groups 2 and 3 because the females in group 1 do not give birth, so y1 = 3x2 + 2x3 . The Group 2 in the following year comes from group 1 only and y2 year comes from group 2 only and y3
= 50% x1 =
= 80% x2 =
4 5
x2 .
1 2
x1 .
The Group 3 in the following
Hence, the populations for the three
groups in the following year can be expressed as ⎛
y1
⎞
⎛
⎜ y2 ⎟ = ⎜ ⎜ ⎝
y3
⎠
0 1 2
⎝ 0
3
2
0
0⎟⎜x ⎟. 2 ⎟ ⎠ ⎝ x3 0⎠
4 5
⎞ ⎛ x1 ⎞
. − →
ii. We write X0 − →
= (x1 , x2 , x3 )
T
− →
.
The vector X0 is called the population distribution. We denote
by Xn the population distribution for the female population in the nth year. Let …
3/7
Solution ⎛ A = ⎜ ⎜ ⎝
0 1 2
3
2
0
0⎟. ⎟
4
0
0
5
⎞
⎠
Then we have − →
− →
− →
− →
− →
− →
2
− →
n
X 1 = AX 0 , X 2 = AX 1 = A X 0 , ⋯ , X n = A X 0 .
− →
⎛
− →
2 X2 = A X0 = ⎜ ⎜
0 1 2
⎝ 0
iii.
3
8
2
5
= ⎜ 0 ⎜
3
⎛
⎝
2 5
2
0
0
⎞
3 0 4 5
⎛
0⎟⎜ ⎟⎜
0 1 2
0⎠⎝ 0
⎞
⎛
3
2
0
0 ⎟ ⎜ 500 ⎟ ⎟
4 5
⎞ ⎛ 1000 ⎞
0⎠
2300
⎝
⎞
1 ⎟ ⎜ 500 ⎟ = ⎜ 850 ⎟ . ⎟ ⎠ ⎝ ⎠ ⎝ 100 400 0⎠
1. Which of the following matrices are row echelon matrices?
4/7
⎞⎛
1000
Section 2.4
…
2
100
⎠
Solution
A = (
1
0
0
0
)B = (
3
0
1
6
⎜0 D = ⎜ ⎜0
0
8
8
0
0
0
⎛
⎝
9
4
3
5
9
0
0
0
7
4
0
⎞
1 ⎛ 6⎟ ⎟E = ⎜0 1⎟ ⎝ 0 ⎠ 0
2
5
0
)C = ⎜0
9
0⎟
0
8
⎛
⎝
0
1
4
1
0
1 ⎟F = ⎜0
0
0⎟
1
0
6
6
⎞
0
0
0
8
1
7
0
0
0
0
G = ⎜2
9
8
0⎟H = ⎜4
0
5
8⎟
0
0
0
9
6
1
⎝
0
⎛
⎠
⎝
0
⎛
⎠
0
⎞
⎠
0
0
⎛
0
⎞
⎝
0
⎞
⎠
2. Which of the following matrices are reduced row echelon matrices? 1
0
0
1
1
5
0
6
0
⎜0 A = ⎜ ⎜0
1
0
0
1
3
0
1
1⎟ ⎜0 ⎟B = ⎜ ⎜0 0⎟
0
0
0
0⎟ ⎟ 1⎟
0
0
0
0
0
0
0
0
1
1
1
1
1
0
0
0
1
1
⎛
⎝
C = (
⎠
⎝
0
⎠
0
1
)D = ⎜0
1
0⎟
0
0
1
0
0
2
0
0
2
1
0
⎝
0
4
0
⎜0 E = ⎜ ⎜0
1
1
3
0
0
0
0⎟ 1 ⎟F = ( 1⎟ 0
0
0
0
0
0
⎞
2
⎛
0
⎝
5/7
⎛
1
⎛
…
⎞
⎞
⎠
⎞
⎠
)
⎞
⎠
Solution ⎛
3. Let A = ⎜ ⎝
1
0
1
2
1
−1 ⎟ .
−3
2
−2
⎞
Use the second row operation to change the numbers 2 and −3 in the first
⎠
column of A to zero. 1
0
4. Let A = ⎜ 0
1
⎛
⎝
0
3
1
⎞
−3 ⎟ . 1
Use the second row operation to change the number 3 in A to 0.
⎠
5. For each of the following matrices, find its row echelon matrix. Clearly mark every leading entry a by using the symbol a and use the symbol ↓ to show the direction of eliminating nonzero entries below leading entries. ⎛ A = ⎜ ⎝
1
0
1
2
1
−1 ⎟ B = ⎜
−3
2
−2
⎞
⎛
⎠
⎝
2
1
0
−1
C = ⎜3
0
3
2
−2
−1
3
⎛
⎝
5
0
1
−2
1
−1
1
−2
1
0
⎞ ⎟ ⎠
⎞ ⎟ ⎠
6. For each of the following matrices, find its reduced row echelon matrix. Clearly mark every leading entry a by using the symbol a and use the symbols ↓ and ↑ to show the directions of eliminating nonzero entries below or above leading entries.
…
6/7
Solution
⎛
2
A = ⎜0 ⎝
⎛
0
0
⎛
⎛
…
7/7
0
1
⎜0 −3 ⎟ B = ⎜ ⎜0 ⎠ −6 ⎝ 0
3
6
−3
0
0
−2
0
0
0
⎞
−2
6
2
0
1
1
0
0
−1
−2
1
1
−2
1
2
2 1
⎜ −3 G = ⎜ ⎜ −2 ⎝
0
1
−3
E = ⎜1 ⎝
1
1
⎛
2
6
C = ⎜2 ⎝
−4
3
0
−6 ⎟ ⎟ 2 ⎟ 2
2
1
0
1
⎟D = ⎜3
0
3
2⎟
−2
−1
3 1
⎞
⎛
⎠
⎝
5
1
1
2
⎟F = ⎜1
1
1
2
1
⎞
⎠
0
1
1
1
2
−2
−1
4
8
1
−3
−7
⎛
⎝ 1
2
⎞
−5 ⎟ ⎟ 0 ⎟ 0
⎠
⎞
⎠
⎞
⎠
⎞
−2 ⎟ 1
⎠
Solution
Solution A, B, C, D are row echelon matrices. A, B, E are reduced row echelon matrices. 1
⎛
3. 3.
A = ⎜ ⎝ ⎛
4. 4.
2
1
1
−3
2
⎛ 1
0
1
⎜ 0 ⎜
1
7
0
−1
0
−3 ⎟ .
1
⎞
−3 ⎟ .
0
10
⎛ 1
0
⎜ ⎜ 0
1
⎝
⎠
1
0
⎠ ⏐ ⎞ ⏐ ⏐R2 (−2)+R3 ⎟ ⏐− → −3 ⎟ ⏐ −−−− ⏐ ⎠ ⏐ 1 ↓ 1
2
⎠
1
1
2
⎞
. −3 ⎟ ⎟
−1
⎜ ⎜ 0
0
1
⎞
1
−1
⎝
⏐ ⎞ ⏐ ⏐R1 (−2)R2 ⏐− − − − − → ⎟ −1 ⏐ ⏐R1 (3)+R3 ⎠ ⏐ −2 ↓
−2
1
⎛
1
1
−2
R2 (−3)(R3 )
⎠
0
⎛ 1
⎝
⎞
1
1
⎝
⎝
1
2
B = ⎜
⎠
R1 (3)+R3
−−−− → ⎜0 −3 ⎟ −
A = ⎜
⎛
1
⎛
R1 (2)+R2
0
3
0
−1 ⎟ − − − − − →⎜0
⎞
1
0
0
0
1
1
⎝
1
−2
⎛
5. 5.
1/9
2 0
0
⎝
…
1
1
−3
A = ⎜0 ⎝
0
0
⎛
1
−1
→ ⎜ ⎟−
0
1
−2
1
⎞
⎠
R1,2
⎝
⏐ ⎞ ⏐ ⏐R2 (1)+R3 ⎟ ⏐− − − − → −2 ⎟ ⏐ − ⏐ ⎠ ⏐ 2 ↓ 1
⏐ ⎞ ⏐ ⏐R1 (2)+R2 ⏐− − − − − → ⎟ −2 ⏐ ⏐ ⎠ ⏐ 0 ↓ 1
t
1. 1. 2. 2.
⎛ 1
−1
⎜ ⎜ 0
1
⎝
0
0
1
⎞
⎟ −2 ⎟ . 0
⎠
Solution 2
1
0
−1
C = ⎜3
0
3
2
⎝
5
−2
R1 (3)+R2
−1
− − − − − →⎜ ⎜ R1 (5)+R3
⎛ 1 ⎜ ⎜ 0
…
2/9
−6
0
3
−6
3
−6
⎟ −3 ⎟ .
0
0
0
0
0
−2
1
2
−2
1
0
0
2
−1
−2
1
0
0
0
0
0
⎛ 1
0
0
⎜ ⎜ 0
1
0
3
0
−1
⎝
0
⎜ ⎜
2
⎝
−3
−3 ⎞
⎝
0
1
−3
E = ⎜ 1
⎝
⎝
1
⎛ 1
⎜ ⎜ 0
⎠
⎛ −1
⎝
⎛ 1
⎝
R2 (−1)+R1
0
−
1 2
0
0
1
−3
3
0
3
5
−2
9
⎠
⏐ −1 ⎞ ⏐ ⎛ 1 ⏐R1 (−1)+R2 ⏐ −−−− → ⎜ 0 1 ⎟ ⏐− ⏐R1 (−2)+R3 ⎠ ⏐ ⎝ 2 0 ↓
↑ −1 ⎞ ⏐ ⏐R3 (2)+R2 ⎟ ⏐ − − − − → −2 ⎟ ⏐− ⏐R3 (1)+R1 ⎠ ⏐ 1 ⏐ 0 ⎞ ⎟ 0 ⎟. ⎠
⏐ −3 ⎞ ⏐ ⏐ ⏐ ⎟ 6 ⏐ ⏐ ⎠ ⏐ 12 ↓
⏐ −3 ⎞ ⏐ ⏐R2 (−1)+R3 ⎟ ⏐− → −3 ⎟ ⏐ −−−− ⏐ ⎠ ⏐ −3 ↓
⏐ −1 ⎞ ⏐ ⏐R2 (1)+R3 ⎟ ⏐ − − − − → −2 ⎟ ⏐− ⏐ ⎠ ⏐ 4 ↓
1
⎛ −1
−−−− → ⎜ ⎟ −
3
⎛ −1
⎞
⎛ 1 ⎜ ⎜ 0 ⎝
0
⎛ 1 ⎜ ⎜ 0 ⎝
0
0
0
−1 ⎞
−2
1
2
−2
1
4
0
2
−1
0
0
0
0
0
−1 ⎞
2
−1
0
0
R2 (−1)
− − → ⎟ − ⎠
R3 (
1 2
)
⎟ → −2 ⎟ −− ⎠
2
0 ⎞
tf
⎛
R2 (
1 2
)
⎟ → 0 ⎟ −− 1
⎠
Solution
⎛
2
−4
A = ⎜0 ⎝
0
0
⎛ 1
−2
⎜ 0 ⎜
1
0
0 1
0
1
⎜0 B = ⎜ ⎜0
3
6
−3
0
0
−2
0
0
0
⎝
0
⎛ 1 ⎜ R4 (1)+R3 ⎜ 0 − − − − − →⎜ R4 (2)+R2 ⎜ ⎜ 0 ⎝
0
⎛ 1 ⎜ ⎜ 0 − −−−− → ⎜ ⎜ ⎜ 0 R2 (−1)+R1
⎝
…
3/9
0
R1 (
1 2
)
↑ 0 ⎞ ⏐ ⏐R2 (2)+R1 ⏐− − − − − → 0 ⎟ ⎟ ⏐ ⏐ ⎠ ⏐ 1 ⏐
1
⎛
⎞
⎛ 1
0
⎞
↑ ⎞ ⏐ ⏐ R3 (3)+R2 ⏐− ⎟ −−−− → −3 ⎟ ⏐ R (−1)+ R1 ⏐ 3 ⎠ ⏐ 1 ⏐
−2
− − − → ⎜ 0 −3 ⎟ − ⎜ 1 ⎠ R3 (− ) 6 ⎝ −6 0
1
6. 6.
⎝
2
1
1 0
⎛ 1
0
0 ⎞
⎜ 0 ⎜
1
. 0 ⎟ ⎟
0
1
⎝
0
⎛ 1 1
0
1
1
2
−1
0
0
1
0
0
0
1
0
1
1
2
−1
0
0
1
0
0
0
1
R2 ( ),R3 (− ) ⎜ 3 2 0 −6 ⎟ ⎜ − − −− − − − → ⎜ ⎟ − 1 ⎜ 2 ⎟ ) R4 ( ⎜ 0 2 ⎠ 2 ⎝ 0
1
⎠
↑ 0 ⎞ ⏐ ⏐ ⎟ ⏐ R (1)+R 0 ⎟ ⏐ 3 2 −−−− → ⎟ ⏐− ⎟ ⏐ 0 ⎟ ⏐R3 (−1)+R1 ⏐ ⎠ ⏐ 1 ⏐
0
−2
0
0 ⎞
1
2
0
0
0
1
⎟ 0 ⎟ ⎟. ⎟ 0 ⎟
0
0
0
1
⎠
⎛ 1 ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎝
0
↑ ⎞ ⏐ ⏐ ⏐ ⎟ −2 ⎟ ⏐ ⎟ ⏐ ⎟ ⏐ −1 ⎟ ⏐ ⏐ ⎠ ⏐ 1 ⏐ 0
1
0
0
1
2
0
0
0
1
0
0
0
↑ 0 ⎞ ⏐ ⏐ ⎟ ⏐ 0 ⎟ ⏐ ⎟ ⏐ ⎟ ⏐ 0 ⎟ ⏐ ⏐ ⎠ ⏐ 1 ⏐
Solution ⎛
0
C = ⎜2 ⎝
2
R1 (−2)+R3
6
−3
−2
6
0
1
⎛ 1
⎛ 1
−1
⎜ 0 ⎜ ⎝
0
2 0
3
R2 (
1 2
)
⎛
)
⎝
↑ ⎞ ⏐ ⏐ R3 (1)+R2 ⏐− ⎟ −−−− → −1 ⎟ ⏐ R (−3)+ R1 3 ⏐ ⎠ ⏐ 1 ⏐ 3
0
0 ⎞
− − − − − →⎜ 0 ⎜
1
. 0 ⎟ ⎟
0
1
0
0
2
−1
−1
3
2
0
1
⏐ ⎞ ⏐ ⏐R2 (−1)+R3 ⏐ ⎟ ⏐− −−−− → −1 ⎟ ⏐ ⎠ ⏐ ⏐ −5 ↓
⎛ 1
⎝
4/9
1
3
2
R2 (1)+R1
…
⎠
2
0
R1 (
− − → ⎜1 ⎟ −
−1
− −−−− → ⎜ 0 ⎜ ⎝
⎞
⎠
⎛ 1 ⎜ 0 ⎜ ⎝
0
⎞
R1 ,2
⎟ −→ ⎜ 0 ⎠
⎛ 1
⎝
−1
2
−1
⎜ 0 ⎜ ⎝
⎛ 1
0 0 ⎞
3
−1 2 0 ⎞
R3 (−
1 4
)
2
− − − → −1 ⎟ − ⎟
0
−4
R2 (
1 2
)
⎠
⎛ 1
2
−− → ⎜ 0 0 ⎟ ⎟ ⎜
0
1
⎠
⏐ ⎞ ⏐ ⏐ ⏐ −1 ⎟ ⏐ ⏐ ⎠ ⏐ 1 ↓ 3
⎝
0
−1 1 0
↑ 0 ⎞ ⏐ ⏐ ⏐ ⎟ 0 ⎟ ⏐ ⏐ ⎠ ⏐ 1 ⏐
Solution 2
1
0
−1
D = ⎜3
0
3
2
⎛
⎝
5
−2
R1 (3)+R2
−1
− − − − − →⎜ ⎜ R1 (5)+R3
⎛ −1 ⎜ ⎜ ⎝
⎝
0
⎝
1
−3
0
3
−6
0
3
−6
−3
−3 ⎞
3
−6
⎟ − − → ⎜ −3 ⎟ − ⎜ 0
0
0
⎠
0 0
1
− − − − − →⎜ ⎜ 0
1
−2
0
0
0
1
−3
3
0
3
5
−2
9
⏐ −3 ⎞ ⏐ ⏐ ⏐ ⎟ 6 ⏐ ⏐ ⎠ ⏐ 12 ↓
⏐ −3 ⎞ ⏐ ⏐R2 (−1)+R3 ⎟ ⏐− → −3 ⎟ ⏐ −−−− ⏐ ⎠ ⏐ −3 ↓
1
⎝
5/9
⎠
⎛ 1
R2 (1)+R1
…
R2 (−1)+R1
R1 (−1)
0
⎛ −1
−−−− → ⎜ ⎟ −
3
⎛ −1
⎞
R2 (
2
1 3
)
⎛ 1
⎝
⎞
⎟ −1 ⎟ . 0
⎠
0
−1
3
1
−2
0
0
↑ ⎞ ⏐ ⏐ ⎟ ⏐ −1 ⎟ ⏐ ⏐ ⎠ ⏐ 0 ⏐ 3
Solution ⎛ 1
0
0
−2
1
2
−2
1
0
0
E = ⎜ 1 ⎝
⎛ 1 ⎜ ⎜ 0 ⎝
0
⎛ 1 ⎜ 0 ⎜
−2
1
0
0 −1
0
0
0
⎛ 1
0
0
⎜ 0 ⎜
1
⎝
6/9
−1
2
⎝
…
2
0
0
−
1 2
0
⏐ −1 ⎞ ⏐ ⎛ 1 ⏐R1 (−1)+R2 ⏐ −−−− → ⎜ 0 1 ⎟ ⏐− ⏐R1 (−2)+R3 ⎠ ⏐ ⎝ 2 0 ↓
⏐ −1 ⎞ ⏐ ⏐R2 (1)+R3 ⎟ ⏐− − − − → −2 ⎟ ⏐ − ⏐ ⎠ ⏐ 4 ↓ ↑ −1 ⎞ ⏐ ⏐R3 (2)+R2 ⏐ − ⎟ − − − → −2 ⎟ ⏐− (1)+R1 R 3 ⏐ ⎠ ⏐ 1 ⏐ 0 ⎞ . 0 ⎟ ⎟ 1
⎠
⎛ 1 ⎜ ⎜ 0 ⎝
0
⎛ 1 ⎜ 0 ⎜ ⎝
0
0
0
0
−2
1
2
−2
1
4
0
2
−1
0
0
0
0
2
−1
0
0
−1 ⎞
−1 ⎞
R2 (−1)
− − → ⎟ − ⎠
R3 (
1 2
)
⎟ → −2 ⎟ −− 2
⎠
0 ⎞
R2 (
1 2
)
−− → 0 ⎟ ⎟ 1
⎠
Solution ⎛ 1
1
2
F = ⎜ 1
1
1
2
1
⎝
⎛ 1 ⎜ ⎜ 0
2
1
2
⏐ ⎞ ⏐ ⎛ 1 ⏐R1 (−1)+R2 ⏐ −−−− → ⎜ 0 −2 ⎟ ⏐− ⏐R1 (−2)+R3 ⎠ ⏐ ⎝ 1 0 ↓ 1
⏐ ⎞ ⏐ ⏐R2 (3)+R3 ⎟ ⏐− − − − → 3 ⎟ ⏐ − ⏐ ⎠ ⏐ −1 ↓ 1
0
1
0
0
−3
⎛ 1
1
2
⎜ 0 ⎜
0
1
0
0
0
⎛ 1
1
0
0 ⎞
⎜ 0 ⎜
0
1
. 0 ⎟ ⎟
0
0
1
⎝
⎝
⎝
0
↑ 1 ⎞ ⏐ ⏐R3 (−3)+R2 ⏐ −−−− ⎟ → 3 ⎟ ⏐− (−1)+R1 R 3 ⏐ ⎠ ⏐ 1 ⏐
⎛ 1
2
0
−1
− − → −3 ⎟ −
0
−3
−1
7/9
⎞
0
1
⎟ → 3 ⎟ −−
0
0
0
8
⎛ 1
1
2
⎜ 0 ⎜
0
1
0
0
⎝
⎝
0
R3 (
1 8
R2 (−1)
⎠
1⎞
)
⎠
↑ 0 ⎞ ⏐ ⏐R2 (−2)+R1 ⏐ −−−− ⎟ → 0 ⎟ ⏐− ⏐ ⎠ ⏐ 1 ⏐
⎠
The last matrix is the reduced row echelon matrix of F.
…
1
2
⎜ ⎜ 0
1
1
Solution
⎜ −3 G = ⎜ ⎜ ⎜ −2 ⎝
3
0
1
1
1
2
−2
−1
4
8
1
−3
−7
0
0
1
1
1
5
1
0
11
11
0
−11
−11
⎛ 1 ⎜ 0 − −−−− → ⎜ ⎜ R2 (−1)+R4 ⎜ 0 R2 (1)+R3
⎝
⎛ 1 ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎝
0
0 1
1
1
5
1
0
1
1
0
−11
−11 1
1
⎜ 0 ⎜ − − − → ⎜ ⎜ ⎜ 0
1
5
1
0
1
1
0
0
0
⎛ 1 ⎜ ⎜ 0 − −−−− → ⎜ R3 (−1)+R1 ⎜ ⎜ 0 R3 (−5)+R2
⎝
0
−5 ⎟ R1 (2)+R3 ⎟ − −−−− → ⎟ 0 ⎟ R1 (−3)+R4 ⎠
⎛ 1
0
1
1
⎜ 0 ⎜ ⎜ ⎜ 0
1
5
1
−1
6
10
1
−6
−10
⎝
1
0
R3 ( ) 11 −2 ⎟ ⎟ − − − → ⎟ ⎟ 0
−1
⎠
⎛ 1 ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎝
↑ ⎞ ⏐ ⏐ ⏐ −2 ⎟ ⏐ R4 (2)+R2 ⎟ −−−− → ⎟ ⏐− ⎟ ⏐R4 (−1)+R1 0 ⎟ ⏐ ⏐ ⎠ ⏐ 1 ⏐ 1
0
0
1
1
0
−4
0
1
1
⎟ 0 ⎟ ⎟. ⎟ 0 ⎟
0
0
0
1
0
0
1
1
1
5
1
0
1
1
⎟ −2 ⎟ ⎟ ⎟ 0 ⎟
0
0
0
−1
…
8/9
1
⎛ 1
0
1
1
⎜ 0 ⎜ ⎜ ⎜ ⎜ 0
1
5
1
0
1
1
0
0
0
⎝
0
0 ⎞
⎠
The last matrix is the reduced row echelon matrix of G.
Section 2.5
⏐ ⎞ ⏐ ⏐ ⏐ ⎟ −2 ⎟ ⏐ ⎟ ⏐ ⏐ 2 ⎟ ⏐ ⏐ ⎠ ⏐ −3 ↓ 1
⎞
⏐ ⎞ ⏐ ⏐ ⎟ ⏐ −2 ⎟ ⏐R3 (11)+R4 − − − − → ⎟ ⏐− ⎟ ⏐ 0 ⎟ ⏐ ⏐ ⎠ ⏐ −1 ↓
0
0
R1 (3)+R2
1
⎛ 1
⎝
⎞
1
0
R4 (−1)
1
⎞
is t
1
⎛
⎠
↑ 0 ⎞ ⏐ ⏐ ⏐ 0 ⎟ ⏐ ⎟ ⎟ ⏐ ⎟ ⏐ 0 ⎟ ⏐ ⏐ ⎠ ⏐ 1 ⏐
Solution
1. For each of the following matrices, find its rank, nullity, and pivot columns. Clearly mark every leading entry a by using the symbol a and use the symbol ↓ to show the direction of eliminating nonzero entries below leading entries.
⎛
1
A = ⎜0 ⎝
⎛ ⎜ D = ⎜ ⎜ ⎝
⎛
0
…
9/9
1
−1
0
0⎟ B = (
0
0
0
1
1
−2
2
1
0 1
⎜0 F = ⎜ ⎜1 ⎝
0
−1
0
−1
0
1
4
2
1
⎜0 ) C = ⎜ ⎜1
2
1
1
0
1
4
0
1
−3
⎛
0
⎞
⎠
0
2
−1
2
−1
3
⎝
⎞
1 ⎛ 0⎟ ⎟ E = ⎜ −2 1⎟ ⎝ 0 ⎠ 0
−1
0
1
1
0
1
0
0
2
0
−1
−1
0
⎞
−2
−1
1
1
0⎟
0
0
1
⎛
2
−2 ⎟ ⎜0 ⎟ G = ⎜ ⎜3 −2 ⎟ 5
⎠
⎝
5
0
2
⎞
⎠
−1
0
−1
3
2
7
0
1
2
−1
1
1
⎞ ⎟ ⎟ ⎟ ⎠
⎞ ⎟ ⎟ ⎟ ⎠
Solution
Solution Because A is a row echelon matrix, r(A) = 2 and null (A) = 4 − 2 = 2. Because columns 1 and 2 of A ⎛
1
⎞
0
⎛
⎞
contain the leading entries, the column vectors ⎜ 0 ⎟ and ⎜ −1 ⎟ are the pivot column vectors of ⎝
0
⎠
⎝
0
⎠
A.
B = (
−1
2
−1
3
R1,2
)− → (
2
−1
3
0
2
−1
∗
) := B .
and null (B) = 3 − 2 = 1. Because the columns 1 and 2 of B∗ contain the leading entries, the
columns 1 and 2 of B : (
0
),(
2
⎛ 1
4
2
⎜ 0 C = ⎜ ⎜ ⎜ 1
2
1
0
1
0
1
⎝
2
2
⎛ 1 ⎜ ⎜ 0 − − − − − →⎜ R2 (4)+R4 ⎜ ⎜ 0 R2 (2)+R3
⎝
0
2
)
are the pivot column of B.
−1
⏐ ⎞ ⏐ ⏐ ⏐ 1 ⎟ ⏐R1 (−1)+R3 ⎟ ⏐− −−−− → ⎟ ⏐ 4 ⎟ ⏐R1 (−2)+R4 ⏐ ⎠ ⏐ −3 ↓ 1
4
2
1
2
1
1
0
1
5
0
1
−1
⎛ 1
4
2
⎜ 0 ⎜ ⎜ ⎜ 0
2
1
⎝
0
⏐ ⎞ ⏐ ⏐ ⏐ ⎟ ⏐R3 (−1)+R4 ⎟ ⏐ −−−− → ⎟ ⏐− ⎟ ⏐ ⎟ ⏐ ⎠ ⏐ ⏐ ↓
⏐ ⎞ ⏐ ⏐ ⏐ ⎟ 1 ⎟ ⏐ ⎟ ⏐ ⏐ 3 ⎟ ⏐ ⏐ ⎠ ⏐ −5 ↓ 1
tf
r(B) = 2
0
−4
−1
−8
−3
⎛ 1
4
2
1
⎞
⎜ 0 ⎜ ⎜ ⎜ 0 ⎜
2
1
1
0
1
5
⎟ ⎟ ∗ ⎟ = C . ⎟ ⎟
0
0
−6
⎝
0
⎠
and null (C) = 4 − 4 = 0. Because the columns 1, 2, 3, 4 of C∗ contain the leading entries, the columns 1, 2, 3, 4 of C are the pivot columns of C. r(C) = 4
…
1/6
Solution ⎛ −1 ⎜ D = ⎜ ⎜ ⎜ ⎝
1
−2
2
1
0
−1
⎛ −1 ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
1
1
⏐ 0⎞ ⏐ ⏐ ⏐R (1)+R 2 ⎟ 0 ⏐ 1 ⎟ ⏐− − − − − → ⎟ ⏐ 1 ⎟ ⏐R1 (2)+R4 ⏐ ⎠ ⏐ 0 ↓
⎛ −1 ⎜ ⎜ ⎜ ⎜ ⎝
1
0
−1
0
3
0
−1
⏐ 0⎞ ⏐ ⏐ ⎟ ⏐ R2 (3)+R3 0⎟ ⏐ −−−− → ⏐− ⎟ ⏐ R2 (−1)+R4 ⎟ 0 ⏐ ⏐ ⎠ ⏐ 0 ↓
0 ⎞
0
−1
0
0
⎟ 0 ⎟ ∗ ⎟ := D . ⎟ 1 ⎟
0
0
0
⎠
and null (D) = 3 − 3 = 0. Because the columns 1, 2, 3 of D∗ contain the leading entries, the columns 1, 2, 3 of D are the pivot columns of D. r(D) = 3
⎛
1
E = ⎜ −2 ⎝
0
−2
−1
1
1
0
0
⏐ 0⎞ ⏐ ⏐R1 (2)+R2 ⏐ − − − − → 0 ⎟ ⏐− ⏐ ⎠ ⏐ ⏐ 1 ↓
⎛ 1
−2
−1
0 ⎞
⎜ 0 ⎜
−3
−1
∗ 0 ⎟ = E . ⎟
0
0
⎝
0
1
⎠
⎛ r(E) = 3
are the pivot columns of E.
2/6
⎞
⎛
and null (E) = 4 − 3 = 1. Because the columns 1, 2, 4 of E∗ contain ⎜ −2 ⎟ , ⎜ ⎝
…
1
0
⎠
⎝
−2 1 0
⎞
⎛
0
⎞
⎟,⎜0⎟ ⎠
⎝
1
⎠
Solution ⎛ 1
−1
0
1
1
0
1
0
0
2
0
−1
−1
⎜ 0 F = ⎜ ⎜ ⎜ 1 ⎝
0
⎛ 1
−1
0
1
1
0
1
0
0
0
0
−1
−1
R2 (−1)+R3 ⎜ 0 − −−−− → ⎜ ⎜ ⎜ 0
⎝
r(F ) = 3
⏐ ⎞ ⏐ ⏐ ⏐R (−1)R 3 ⎟ −2 ⏐ 1 ⎟ ⏐− − − − − → ⎟ ⏐ −2 ⎟ ⏐ ⏐ ⎠ ⏐ 5 ↓ 0
0
1
⎞
⎛
⎜0⎟ ⎜ ⎟,⎜ ⎜1⎟ ⎜
columns 1, 2, 3 of F: ⎜ ⎝
2
⎛ −1 ⎜ ⎜ ⎜ ⎜ ⎝
…
3/6
5
0
−1
3
2
7
0
1
2
−1
1
−1
−1
0
3
2
0
−3
−2
0
0
−1
⎜0 G = ⎜ ⎜3 ⎝
−1
0
1
⎜ 0 ⎜ ⎜ ⎜ 0
1
0
1
1
0
1
0
−1
−1
⎝
0
⎞
⎛ 1
−1
0
1
R3,4 −2 ⎟ ⎟− → ⎟ 0 ⎟
⎜ 0 ⎜ ⎜ ⎜ ⎜ 0
1
0
1
0
5
⎠
⎝
0
0
−1
−1
0
0
0
⏐ ⎞ ⏐ ⏐ ⏐ ⎟ −2 ⎟ ⏐ ⎟ ⏐ ⏐ −2 ⎟ ⏐ ⏐ ⎠ ⏐ 5 ↓ 0
0
⎞
−2 ⎟ ⎟ ⎟ = F ⎟ 5 ⎟ 0
∗
.
⎠
ansd null (3) = 5 − 3 = 2. Because the columns 1, 2, 3 of F* contain the leading entries, the ⎛
⎛
⎛ 1
−6
−4
7
⎠
⎝
−1 1 0 0
⎞
⎛
⎟ ⎜ ⎟,⎜ ⎟ ⎜ ⎠
⎝
0 0 0 −1
⎞
⎛ −1
⎟ R3 (−1)+R1 −−−− → ⎟ − ⎟
⎜ ⎜ ⎜ ⎜
⎠
⎝
3
2
7
3
0
1
2
5
−1
1
1
−1
−1
0
3
2
7
0
0
0
0
0
0
0
0
⎜ ⎜ ⎜ ⎜
−14
⎠
0
⎟ R2 (1)+R3 ⎟ − − − − − → ⎟ R2 (2)+R4 ⎟ −7
⎝
are the pivot columns of F.
−1
⎛ −1
⎠
⎟ ⎟ ⎟
−1
−3 ⎞ 7
⎞
−3 ⎞ ⎟ R1 (3)+R3 ⎟ − − − − − → ⎟ ⎟ R1 (5)+R4 ⎠
−3 ⎞ ⎟ ⎟ = G∗ . ⎟ ⎟ ⎠
Solution r(G) = 2
and null (G) = 4 − 2 = 2. Because the columns 1, 2 of G* contain the leading entries, the ⎛
2
⎞
⎛
⎜0⎟ ⎜ ⎟,⎜ ⎜3⎟ ⎜
columns 1, 2, 4 of G: ⎜ ⎝
5
⎠
⎝
−1 3 0 −1
⎞ ⎟ ⎟ ⎟
are the pivot columns of G.
⎠
Section 2.6 1. Determine which of the following matrices are elementary matrices. 1. 2. 3. 4. 5. 6.
(
(
(
(
2
0
0
1
0
1
1
0
1
5
0
1
2
0
0
2
0
2
2
0
(
0
−2
1
) 0
0
1
⎜0
1
0⎟
0
0
1 1
⎜ −1 ⎝
4/6
)
1
⎛
…
)
(
⎝
8.
)
)
⎛
7.
)
0
⎞
⎠
0
0
1
0⎟
0
1
⎞
⎠
Solution 1
⎛
9.
⎜ −1 ⎝
0⎟
0
1
⎠
0
0
⎜0 ⎜ ⎜0
1
0
0
5
0⎟ ⎟ 0⎟
0
0
0
1
1
0
0
0
⎜0 ⎜ ⎜0
1
0
0
0
0⎟ ⎟ 0⎟
0
0
0
1
1
0
0
0
⎜0 ⎜ ⎜0
0
0
0
1
1⎟ ⎟ 0⎟
1
0
0
⎛
⎝ ⎛
12.
1
⎞
0
⎝
11.
0
1
⎛
10.
1
0
⎝
0
⎞
⎠ ⎞
⎠ ⎞
⎠
2. Use row operations to change the above matrices (1), (2), (3), (6), (7), (8), (10), and (12) to an identity matrix. 3. Let I
= (
1
0
0
1
).
1.
E2 (5)E2 (
2.
E1(2)+2 E
3.
E1,
2 E1, 2
1(
1 5 1 2
Verify the following assertions. ) = I; )+2
= I;
= I.
4. Find a matrix E such that B = EA, where
…
5/6
Solution ⎛
0
A = ⎜1 ⎝
…
6/6
2
2
−1
−1
3
0
1
⎞ ⎟ ⎠
⎛
1
and B = ⎜ 0 ⎝
0
−1
3
⎞
2
−1 ⎟ .
0
−4
⎠
Solution
Solution 1. 1. 1.
(
2
0
0
1
)
is an elementary matrix because
(
2. 3.
0
1
(
(
) 1
0
1
5
0
1
(
2
0
0
2
2×2
0
0
1
is an elementary matrix because (
R1 (2)
) − − → (
1
)
)
−1
0
0
1
R1,2
0
).
0
1
1
0
1
5
0
1
)− → ( 0
1
).
is an elementary matrix because
(
4.
1
1
0
0
1
R2 (5)+R1
) − − − − − →(
).
is not an elementary matrix because we need to use two row operations to change the
identity matrix I to the matrix: (
1
0
0
1
R1 (2)
) − − → (
2
0
0
1
R2 (2)
) − − → (
2
0
0
2
).
(6), (7), (8), (10), (12) are elementary matrices, but (5), (9), and (11) are not elementary matrices. 2. 2.
…
1/6
Solution
1. 2. 3.
6.
(
(
(
2
0
0
1
0
1
1
0
1
5
0
1
1
R1 (
1 2
)
) − − − → (
R1 ,2
) −→ (
1
0
0
1
1
0
0
1
).
R2 (−5)+R1
) − −−−− → (
0
(
).
1
0
0
1
R1 (2)+R2
).
1
0
0
1
) − − − − − →( −2
1
).
The row operations used in (7), (8), (10), (12) are R1,3 , R1 (1) + R2 , R
1
, R2,4 ,
5
3. 3.
Because 0
(
R2 (5)
1
0
) − − → ( 0
1
1 ) = E2 (5),
0
(
5
E2 (5)E2 (
5
) = (
1
0
0
5
R2 (
1 5
)
1
) − − − → ( 0
1
0
1 )( 0
0
1 0 1
) = (
5
1
0
0
1
0 ) = E2 (
is
1
1
1 5
),
5
) = I.
Because 1
0
(
R1 (2)+R2
1
0
2
1
) − − − − − →( 0
1
) = E1(2)+2
and (
we have
…
2/6
1
0
0
1
R1 (−2)+R2
) − −−−− → (
1
0
−2
1
) = E
1(
1 2
)+2,
respectively.
Solution 1 E1(2)+2 E
1(
1
)+2
2
0
= (
1
0
)( 2
1
1
0
0
1
) = ( −2
) = I.
1
Because (
1
0
0
1
E1,2 E1,2 = (
⎛ A =
0
⎜1 ⎝
4. 4. ⎛
2
1
⎜0 ⎝
0
2
−1
−1
3
0
1
−1 2 2
3
⎞
R1,2
⎛
1
⎠
⎞
⎝
2
1
0
3
R2 (1−)+R3
⎛
−5
⎝
1
⎝
⎛
1
I3 = ⎜ 0 ⎝
⎛
0
1
0
0
1
1
0
) = E1,2 ,
) = (
3
0
0
1
) = I.
⎞
2
−1 ⎟ = B.
0
−4
⎠
0
0
0
1
0
1
→ ⎜1 0⎟−
0
0 ⎟ = E1 .
0
1
0
1
0
⎞
⎞
⎠
R1,2
⎛
⎝
R1 (−2)+R3
0
⎛
−−−− → ⎜ 0⎟ −
0
0
1
1
0
0
0
1
R1 (−2)+R3
1
I3 = ⎜ 0 ⎝
0
1
⎠
−1
0
1
⎞
1
−−−− → ⎜0 −1 ⎟ − ⎠
)(
0
−−−− → −1 ⎟ −
0
I3 = ⎜ 0
3/6
1
2
⎛
…
0
−1
→ ⎜0 ⎟−
R1,2
)− → (
⎠
⎝
⎞
⎠
1
0
0
0
1
0 ⎟ = E2 .
−2
0
1
⎞
⎠
1
0
0
1
−−−− → ⎜0 0⎟ −
1
0 ⎟ = E3 .
0
1
⎞
⎠
R2 (−1)+R3
⎛
⎝
0
−1
1
⎞
⎠
Solution
Let E
= E3 E2 E1 .
Then 1
0
0
E = ⎜0
1
0⎟⎜
⎛
⎝
0
−1
1
1
0
0
0
1
0
0
1
0⎟⎜1
0
0⎟
−2
0
1
0
1
⎞⎛
⎠⎝
1
0
0
= ⎢⎜ 0
1
0⎟⎜
⎡⎛
⎣⎝
⎛ = ⎜ ⎝
−1
1
0
0
1 0
1
−2
⎠⎝
−1
1
1
0
1
0
0⎟⎜1
−1
−2
EA = ⎜
1
⎞⎛
⎠⎝
0
2
0
⎠
0
0
0
1
0
0
1
0 ⎟⎥ ⎜ 1
0
0⎟
0
1
0
0⎟⎜1 ⎠⎝
⎠⎝
⎞
1
−2
⎞⎛
0
⎛
⎝
0
⎞⎛
⎞⎛
0
0
1
⎞⎤ ⎛
⎠⎦ ⎝
1
0
0
0⎟ = ⎜
0
1
2
−1
−1
3
0
1
⎞
⎛
⎠
⎝
⎞
⎛
0
⎝
⎠
0
1
0
1
0
0⎟
−1
−2
1
−1
⎟ = ⎜0 ⎠
⎞
0
1
3
⎞
⎠
⎞
2
−1 ⎟ = B.
0
−4
⎠
Section 2.7 1. Show that B is an inverse of A if
A = (
2. Use Definition 2.7.1
to show that (
1
1
−1
1
1
−1
−2
2
1
)
)
and B = (
2
−
1
1
2
2
1 2
).
is not invertible.
3. For each of the following matrices, determine whether it is invertible. If so, find its inverse.
…
4/6
Solution 1
⎛
3
A = ⎜ 0 ⎜ ⎝
C = (
4. Let A = ( 5. Let A = ( −1
(AB)
0
0
0⎞
1
⎟ 0⎟
4
0
4
3
−4
2
3
)
1
0
0
0
⎜0 B = ⎜ ⎜0
2
0
0
3
0⎟ ⎟ 0⎟
0
0
4
⎛
⎠
⎝
0
2
4
−1
−2
D = (
⎞
⎠
)
2
1
x
1
1
1
2
1
3 −1
= B
Find all x ∈ R such that A is invertible.
).
and B = (
)
−1
A
3
2
2
2
).
−1
Find A−1 , B−1 , and (AB)
and verify
.
6. For each of the following matrices, use its rank to determine whether it is invertible.
⎛
0
2
A = ⎜1 ⎝
5
1
−5
⎛ ⎞
−5 ⎟
0
8
⎠
1
⎜ 2 B = ⎜ ⎜ −2 ⎝
1
0
0
0
1
2
0
−3
1
−1
⎞
⎟ ⎟ −2 ⎟
0
0
⎛ C = ⎜ ⎝
⎠
2
1
4
1
−2
4
−2
−1
−4
⎞ ⎟ ⎠
7. For each of the following matrices, use row operations to determine if it is invertible. If so, find its inverse. A = (
−1
2
4
7
⎛
1
D = ⎜0 ⎝
…
5/6
2
−1
)
B = (
1
⎞
1
−1 ⎟
0
−1
⎠
1
−3
−4
5
)
1
1
E = ⎜2
0
⎛
⎝
3
2
C = (
2
⎞
−5 ⎟ 1
⎠
1
−2
−2
4 ⎛
)
0
F = ⎜2 ⎝
1
−8 4 −2
−9
⎞
−1 ⎟ −5
⎠
Solution
8. Let A = ( 9. Let A = (
−1
2
1
3
−4
1
3
10. Let A = (
1
6/6
Find A3 , A−1 , (A−1 )
).
Find A−1 , (AT )
−1
3
,
and verify (A3 )
and verify that (AT )
1 3 ).
3
…
).
2
Find A−1 . Is A−1 symmetric?
−1
−1
−1
= (A
−1
= (A
3
) .
)
T
.
Solution
Solution 1. 1.
Because 1
1
−1
1
AB = (
1
−
2
)(
1
1
2
2
1 2
) = (
1
0
0
1
) = I2 ,
by Definition 2.7.1, B is the inverse of A.
1
−1
−2
2
b12
b21
b22
)(
)
b11
b12
b21
b22
by Definition 2.7.1, ( 3. 3.
be a 2 × 2 matrix. Because b11 − b21
b12 − b22
−2b11 + 2b21
−2b12 + 2b22
) = (
1
−1
−2
2
)
⎛ A
∣3
Because |C| = ∣
∣2
−4 ∣ ∣ ≠ 17 ≡ 0, 3
3
= ⎜0 ⎝
0
0
1
),
0
0 4
0
⎛ ⎞ −1
0 ⎟,B
0
1
⎠
4
1
0 1
⎜0 ⎜ = ⎜ ⎜0 ⎜ ⎝
2
0
0
0
0
0
0
0 ⎟ ⎟ ⎟. 0 ⎟ ⎟
1 3
0
1 4
so C is invertible and
∣
C
−1
1 =
3
4
−2
3
( 17
1/7
1
Because A and B are diagonal matrices, by Example (2.7.2),
−1
…
) ≠ (
is not invertible.
t
(
b11
r
Let (
2. 2.
) = (
3
4
17
17
−
2
3
17
17
).
⎞
⎠
Solution
4. 4.
∣ ∣ = 0, −2 ∣ 4
so D is not invertible. 2
∣1
Because |A| = ∣
2
∣ = 1−x 1 ∣
∣1
∣1
2∣ −1 = ∣ = 1, A 3∣
Because |A| = ∣
∣1 ∣3
2∣ −1 = ∣ = 2, B
∣2
2∣
Because |B| = ∣
Then x = −1 or x = 1. Hence when x ≠ 1 and
= 0.
and A is invertible.
2 x ≠ 1, ∣ ∣A∣ ∣ = 1−x ≠ 0
5. 5.
∣
x
−1
B
−1
A
1
|A|
−2
−2
3
1
−1
= (
1
2
1
3
)(
3
2
2
2
) = (
7 9
−1
(AB)
=
−1
⎛
0
A = ⎜1 ⎝
1
R3 (−2)+R2
2
−5
0
8 1
− −−−− → ⎜0 ⎝ r(A) = 3
…
2/7
⎞
−1
A
R1,3
⎛
0
0
⎠
8
⎝
1
0
0
8
1
−1
1
8
−6
−9
7
).
2
−2
∣7 ) , |AB| = ∣ ∣9 8
(
).
3
3
6
4 ) = ( −
4 −
−−−− → ⎜0 −5 ⎟ −
5
2
−5
⎠
⎝
1
− − − − →⎜0 −3 ⎟ −
1
2
−5
⎛
⎝
0
⎛
0
8
⎞
−3 ⎟ , −11
⎠
0
2
−3
9
7
2
2
6∣ ∣ = 2, 8∣
) = (
5
R1 (−1)+R2
0
and A is invertible.
)(
0
R2 (2)+R3
⎠
−1 −1
−1
1
⎞
1
⎞
−2
8
7
2
2
⎞
−13 ⎟ −5
−3
9
.
→ ⎜1 −5 ⎟ −
5
⎛
−1
= B
1
2
2
Hence, (AB) Because 6. 6.
a
) = (
3
1
3 ) = (
−c
−1
Because AB = (
−b
(
2
(
2
d
1
ri
∣ 2 |D| = ∣ ∣ −1
⎠
).
).
Solution 1
2
−2
⎜0 = ⎜ ⎜0
1
−3
2
1
0
−2
⎛ T
B
⎝
0
1
1
2
−2
−1 ⎟ R2 (−2)+R3 ⎜ 0 −−−− → ⎜ ⎟ − ⎜0 0 ⎟
1
−3
0
7
0
−2
2
−2
R3 ( )+R4 ⎜0 7 − − − − − → ⎜ ⎜ ⎜0
1
−3
0
7
0
0
⎝0
⎛
⎠
0
1
⎛ 2
⎞
⎝
1
0
1
⎞
−1 ⎟ ⎟ 2 ⎟ ⎠
0
⎞
−1 ⎟ ⎟. ⎟ 2 ⎟ 4
⎠
7
This implies that r(B) = r(BT ) = 4 and B is invertible. Because ⎛ C = ⎜ ⎝
2
1
4
⎞
−2
4
−2
−1
−4
⎛
1
−2
− −−−− → ⎜0 ⎝ r(C) = 2
⎠
4
2
⎝
1
4
⎛
R1,2
1
−2
→ ⎜2 4⎟−
−2
0
⎞
0
0
⎠
⎝
0
4
0
0
4⎟
0
0
⎞
⎠
and C is not invertible.
7. 7. (A|I) = (
R1 (−1)
1
−1
2
1
0
4
7
0
1
−2
−1
− −− → ( R2 (
1 15
)
0
Hence, A−1
…
3/7
1 − = (
R1 (4)+R2
) − − − − − →(
0
R2 (2)+R1
1
15
15 2
15
15
4
1
15
15
2
1
0
0
15
4
1
⎛
1
0
⎝
0
1
) − − − − − →
4
7
−1
).
−
)
7
2
15
15
4
1
15
15
⎞
1
−4 ⎟ ,
5
0
⎛
− − − − →⎜1 ⎟ −
1
R1 (−2)+R2
R1 (1)+R3
⎞
. ⎠
⎠
Solution ∣ ∣ (B I) = ( ∣ ∣ R2 (−
1 7
1
−3
1
0
−4
5
0
1
1
)
−3
1
−−− → ( 0
1
− −
Hence, B−1
= ( −
7
−
7 4
−
7
7 1
4
0
1
)
(D|I) = ⎜ ⎜
0
1
−1
0
1
⎝
2
0
−1
0
0
1
R2 (−2)+R3
1
1
⎛
− −−−− → ⎜ ⎜ ⎝
⎜ ⎜
0
⎝
0
0
1 0
Hence, D
…
4/7
−
⎝
0
1
−
)
5
−
7 4
−
7
1
−2
1
0
0
0
2
1
3 7 1 7
⎞
. ⎠
).
1
−1
1
1
0
0
⎞
0
⎟ − → ⎜ ⎜ ⎟ −−−−
0
1
−1
0
1
0
⎟ ⎟
1
⎠
⎝
0
2
−3
−2
0
1
⎠
−1
R1 (−2)+R3
1
1
0
0
⎞
⎟ − −−−− → ⎟
0
0
0
−1
−2
−2
1
−2 0
⎛
⎞
1
3
−2
0
0
−2
2
0
−1
= ⎜2
3
−1 ⎟
2
−1
2
0
1
1
⎝
1
0
1
⎛ −1
−1
0 −1
4
contains a row of zeros, the matrix C is not invertible.
−1
−1
−7
1
R1 (2)+R2
1
1
0
⎛
) − − − − − →(
⎛
⎛
0
7
−2
0
1
).
0
0
−3
3
1
−2
1
) − − − − − →
7
−2
1
R2 (3)+R1
1
−
1
(C|I) = (
Because (
0
4
5
R1 (4)R2
) − − − − → (
⎞
⎠
1 −1 1
⎠
R3 (1)+R1
R3 (−1)+R2
⎛
1
0
0
1
1
0
⎟ − − − − →⎜ ⎜ ⎟ −
0
1
0
2
3
−1
⎟. ⎟
⎝
0
0
1
2
2
−1
⎠
⎞
⎠
R2 (1)+R1
R3 (−1)
⎞
1
(E|I) = ⎜ ⎜
2
0
−5
0
1
⎝
3
2
1
0
0
2
1
0
0
⎛
1
1
2
1
0
0
⎞
0
⎟ − → ⎜ ⎜ ⎟ −−−−
0
−2
9
−2
1
0
⎟ ⎟
1
⎠
⎝
0
−1
−5
−3
0
1
⎠
− → ⎜ ⎜
0
−1
−5
−3
0
1
0
−2
9
−2
1
0
0
⎞
R1 (3)+R3
R1 (−2)+R2
⎛
1
1
2
1
0
0
⎞
⎟ − − − → ⎜ ⎜ ⎟
0
1
5
3
0
−1
⎟ ⎟
0
−2
9
−2
1
0
⎞
R2 (−1)
⎝
⎠
⎛
1
1
2
1
0
0
− − − − − →⎜ ⎜
0
1
5
3
0
−1
⎟ − → ⎟ −−−−
⎝
0
0
1
4
1
−2
⎠
0
−7
−2
⎜ ⎜
0
1
0
−17
−5
9
⎝
0
0
1
4
1
−2
10
⎛
Hence, E
−1
4
= ⎜ −17 ⎝
−8
⎛
1
0
0
10
3
−5
⎟ − → ⎜ ⎜ ⎟ −−−−
0
1
0
−17
−5
9
⎝
0
0
1
4
1
−2
⎞
−5
−5
9
1
−2
−9
⎛
0
(F |I) = ⎜ ⎜
2
4
−1
0
1
⎝
1
−2
−5
0
0
1
0
⎛
1
−2
−5
0
0
− −−−− → ⎜ ⎜
0
8
9
0
1
0
−8
−9
1
0
R1 (−2)+R2
⎝
⎛
1
−2
−5
0
0
1
⎜ ⎜
0
8
9
0
1
−2
⎟. ⎟
0
0
0
1
1
−2
⎠
⎝
R2 (−1)+R1
⎠
3
4
⎠
⎞
rD
1
R3 (−5)+R2
⎞ ⎟ ⎠ 0
fo
1
R3 (−2)+R1
⎛
1
−2
−5
0
0
1
⎞
0
⎜ ⎟− ⎟ → ⎜
2
4
−1
0
1
0
⎟ ⎟
1
⎠
⎝
0
−9
1
0
0
⎠
⎞
R1,3
ot
⎛
⎞
N
R2 (2)+R3
5/7
0
1
⎝
…
1
⎛
R2,3
1
2
rib ut io
1
1
−2 0
⎞
−8
R2 (1)+R3
⎟ − − − − − → ⎟ ⎠
⎞
⎟. ⎟
is t
⎛
n
Solution
⎠
Solution
Because the left of the last matrix contains a row of zeros, F is not invertible. 8. 8. 2
2
= A
3
⋅A = (
1
3
4
2
11
∣ −1
4
3
2
11
)(
−1
2
1
3
)
) = (
1
18
9
37
1
−1
A
1
=
3
(
−5
−1
(A
)
−
2
= (
(A
)
3
−1
= (A
2
−1
) A
3
2
5
5
1
1
5
5
)(
3
(A )
9. 9.
−1
∣ −4 |A| = ∣ ∣
…
6/7
3
−1
= (A
= −
5
1
1
5
5
2
5
5
1
1
5
5
25
) = (
4
25
).
11
−
−
)(
25
3
2
5
5
1
1
5
5
−
4
25
2
3
25
25
− ) = (
37
−18
−9
1
(
−
).
37
18
125
125
9 125
) = (
37
18
125
125
9 125
3
) .
1∣ −1 = − ∣ = −7; A 1∣
2
5
3
25
1
−1
125
Hence, (A3 )
3
3
ot
and
N
18 ∣ ∣ = −125 37 ∣
) = (
fo
−
∣1 3 |A | = ∣ ∣9
−
2
−
−1
−
25
= (
−2
−1
11 −1
)
2∣ ∣ = −5, 3∣
Because |A| = ∣ ∣
3 ) = (
n
3
2
rib ut io
1
A
−1 )(
is t
= (
rD
−1
2
A
1 7
(
1
−1
−3
−4
− ) = (
1
1
7
7
3
4
7
7
)
−
1 125
).
−
1 125
).
Solution
T
A
= (
−4 1
−1
(A
10. 10.
…
7/7
)
3
) , (A
= (
−1
A
)
−1
= −
1 −
T
T
=
1
3
7
7
1
4
7
7
1 |A|
(
).
2 −3
1 7
1
(
−1
Hence, (AT ) −3 1
) = −
−3
1 7
) = (
−4
−1
(
−
−1
= (A
2 −3
−3 1
)
T
1
3
7
7
1
4
7
7
and
)
.
− ) = (
3 7
2
3
7
7
−
−1
1 7
),A
is symmetric.
A.3 Determinants
A.3 Determinants Section 3.1 a
b
c
d
),
where a, b, c, d ∈ R. Show that
io n
1. Let A = (
bu t
2 ∣ ∣λI − A∣ ∣ = λ − tr(A)λ + ∣ ∣A∣ ∣
2. Let A1
2
4
= (
1 ),
3
2
A2 = (
−2
), −1
rD is tri
for each λ ∈ R. −1
1
−1
−3
A3 = (
4
For each i = 1, 2, 3, find all
)
such that |λI − Ai | = 0. and (3.1.4) to calculate each of the following determinants. 3. Use (3.1.3) λ ∈ R
|A| =
∣ ∣
2
3∣
−2
1
2
3
−1
∣ ∣
4∣
∣2
−2
|B| =
∣
4∣ ∣
4
3
1
∣0
1
2∣
∣
∣0
|C| =
∣
ot
∣
1
fo
∣
∣
−1
3∣ ∣
4
1
2
∣0
0
1∣
∣
∣
N
4. Compute the determinant |A| by using its expansion of cofactors corresponding to each of its rows, where ∣ |A| =
∣ ∣ ∣
5. Compute each of the following determinants. …
1/2
1
2
3∣
−2
1
2
3
−1
∣ ∣
4∣
.
A.3 Determinants ∣2 |A| =
∣ ∣
0
∣0
…
2/2
3 6 0
−4 ∣ 0 5
∣ ∣ ∣
∣ −2 |B| =
∣ ∣ ∣
0
0∣
1
1
0
0
3
8∣
∣ ∣
Solution
Solution Because λI − A = λ (
1
0
0
1
)−(
a
b
c
d
) = (
λ−a
−b
−c
λ−d
2
= λ
2. 2.
∣2
Because |A1 | = ∣
∣3
4
2
− (a + d)λ + (ad − bc) = λ
∣ ∣ = −16
2
2
ot
and tr(A1 ) = 5, 2
|λI − A2 | = λ
N
∣ −1
2∣ ∣ = 6 4∣
− 16 = (λ − 4)(λ + 4) = 0.
fo
− tr(A1 )λ + |A1 | = λ
Hence, λ = 4 or λ = −4. ∣1
− tr(A)λ + ∣ ∣A∣ ∣.
and tr(A1 ) = 2 + (−2) = 0,
−2 ∣
|λI − A1 | = λ
Because |A2 | = ∣
∣ ∣ = (λ − a)(λ − d) + bc λ−d∣ −b
rD is tri
∣λ − a |λI − A| = ∣ ∣ −c
bu t
we have
),
io n
1. 1.
2
− tr(A2 )λ + |A2 | = λ
− 5λ + 6 = (λ − 2)(λ − 3) = 0.
Hence, λ = 2 or λ = 3. ∣ −1
Because |A3 | = ∣
∣ −1
∣ ∣ = 4 −3 ∣ 1
and tr(A1 ) = −4, 2
|λI − A3 | = λ
…
1/5
2
− tr(A3 )λ + |A3 | = λ
+ 4λ + 4 = (λ + 2)
2
= 0.
Solution
Hence, λ = −2.
∣ ∣
1 −2
1
3
−1
∣2
3. 3.
|B| =
|C| =
∣
3∣
2
−2
2
∣ ∣
4∣ 4∣ 2 ∣
1
2
−2
1
3
−1
−2 = (12 + 0 + 16) − (0 + 2 − 16) = 42.
4
3
1
4
3
∣0
1
2∣ 0
1
∣0
−1
3∣ 0
−1
∣
∣
∣
∣
4
1
2
4
1
∣0
0
1∣ 0
0
∣
∣
= (4 + 12 + 6) − (9 − 2 − 16) = 31.
n
|A| =
∣
is tri bu tio
∣
= (0 + 0 + 0) − (0 + 0 − 4) = 4.
4. 4.
∣ −2 M12 = ∣ ∣ 3
2∣ 1+2 3 M12 = (−1) M12 = −M12 = −(−14) = 14. ∣ = −14. A12 = (−1) 4∣
∣ −2 M13 = ∣ ∣ 3
∣ 1+3 4 M13 = (−1) M13 = M13 = −1. ∣ = −1. A13 = (−1) −1 ∣
rD
2∣ 1+1 2 M11 = (−1) M11 = M11 = 6. ∣ = 6. A11 = (−1) 4∣
fo
1
ot
1.
∣ 1 M11 = ∣ ∣ −1
|A| = a11 A11 + a12 A12 + a13 A13 = A11 + 2A12 + 3A13
N
= 6 + 2(14) + 3(−1) = 31.
…
2/5
Solution ∣ 2 M21 = ∣ ∣ −1 ∣1 M22 = ∣ ∣3
3∣ 2+2 M22 = M22 = −5. ∣ = 4 − 9 = −5. A22 = (−1) 4∣
∣1 M23 = ∣ ∣3
∣ 2+3 M13 = −M23 = 7. ∣ = −1 − 6 = −7. A23 = (−1) −1 ∣ 2
io n
2.
3∣ 2+1 M21 = −11. ∣ = 8 − (−3) = 11. A11 = (−1) 4∣
|A| = a21 A21 + a22 A22 + a23 A23 = −2(−11) + (−5) + 2(7) = 31.
bu t
3∣ 3+1 M31 = M31 = 1. ∣ = 4 − 3 = 1. A31 = (−1) 2∣
∣ 1 M32 = ∣ ∣ −2
3∣ 3+2 M32 = −M32 = −8. ∣ = 2 − (−6) = 8. A32 = (−1) 2∣
is tri
3.
∣2 M31 = ∣ ∣1
∣ 1 M33 = ∣ ∣ −2
rD
2∣ 3+3 M33 = M33 = 5. ∣ = −1 − (−4) = 5. A33 = (−1) 1∣
By Theorem 3.1.3, |A| = 2(6)(5) = 60; |B| = (−2)(1)(8) = −16.
ot
5. 5.
fo
|A| = a31 A31 + a32 A32 + a33 A33 = 3(1) − (−8) + 4(5) = 31.
Section 3.2
N
1. For each of the following matrices, find M32 , M24 , A32 , A24 , M41 , M43 , A41 , A43 ⎛ ⎜ A = ⎜ ⎜ ⎝
…
3/5
1
−1
2
3
2
2
0
3
1
2
3
−3
0
1
⎞
⎛
⎟ ⎟ −1 ⎟
⎜ B = ⎜ ⎜
⎠
⎝
6
1
0
2
−1
2
0
0
1
1
0
1
−1
0
1
⎞
⎟ ⎟. −1 ⎟ 2
⎠
Solution
2. Compute each of the following determinants by using its expansion of cofactors of each row.
∣ |A| =
∣ ∣ ∣
1
0
2
2
0
0
1
∣ −1
−1 ∣
∣
∣
∣
∣
∣
1
0
1
−1
0
1
3
|B| =
∣
∣
−1
2
3
2
2
0
3
1
2
3
−1
0
1
6
∣
∣
∣
∣ −3
0
0
1∣
1
0
∣ 1∣
0
0
1
0
∣1
1
−1
|A| =
∣0 ∣ ∣
0
2
0
∣0
0
6∣
∣1
−1
∣ ∣
∣ ∣
0
∣ ∣
5
∣2 ∣
1
∣ ∣
…
4/5
∣
∣ ∣
2
∣0
3
0
0
0
−2 1
0
6
3
−4
−2
6
10 0
2
0
0
−1
6
−6
0
4
0
2
∣ ∣
∣
|C| =
3
0
1
2
∣0
0
∣ ∣
∣ ∣ ∣
−1 ∣ 5
0∣
0
3
0
0
0
|E| = ∣ 0
0
−1
2
7∣
0
0
0
−4
∣0
0
0
0
∣
∣ −2
0
0
0∣
∣
∣
∣
∣
0
3
0
∣ 0∣
0
0
−1
0
0
0
6
∣1
2
3
7∣
2
.
−1
0
1
∣
∣1
2∣
∣
0
0
∣
0∣
0∣
0∣
5
0
∣ −2 ∣
|B| =
0
∣ |F | =
∣
N
|D| = ∣ 0
∣2
fo
0∣
ot
|A| =
0
∣
rD
4. Evaluate each of the following determinants by inspection. ∣1
.
∣
∣
is tri
∣
∣
bu t
3. Evaluate |A| by using the expansion of cofactors of a suitable row, where ∣0
∣
1
io n
∣
∣
∣
∣
∣
∣
∣
∣
0∣
∣
|G| =
∣
∣
∣
−1 ∣
∣
0
∣ ∣
2∣
6
∣ ∣ ∣ ∣
2∣
Solution
…
5/5
9/9/2020
Solution
Solution
∣ −3
1 6∣
∣
−1 2 ∣
∣ ∣
1 1
∣ −3 A32 = (−1) ∣ −1 M41 =
∣ ∣ ∣
∣
2
3 −1 ∣
A41 = (−1)
M32 =
∣
1 2
∣ −1 A32 = (−1) ∣ M24 =
∣ ∣
1 1
∣ −1 A24 = (−1)
…
1/9
∣
3
2 4+1
−1
1
2
−3
0
3 3
−1
∣ ∣
∣ ∣ ∣
−1 ∣
2+4
0 = (0 + 12 + 18) − (0 − 9 − 4) = 43.
2
3 −1
2
2
1
2
1
1
2
∣ ∣ ∣
M24 = M24 = 24.
2
= (−2 − 3 + 12) − (6 + 6 + 2) = −7.
M41 = −M41 = −43. A43 = (−1)
0
4+3
M43 = −M43 = 7.
1
2
2
0 = (0 − 2 − 2) − (0 + 1 + 8) = −13.
−1
1
M32 = −M32 = 13.
0 2∣ 0 1
∣ ∣
0 1∣ 2+4
2
1
2 −1 ∣
3+2
= (2 + 9 + 0) − (−12 + 0 − 1) = −24.
M32 = −M32 = 39. A24 = (−1)
0
∣1
∣
1∣
2
2
∣
0
1
1
∣
3
2
∣
3
−1
0 = (0 − 18 + 6) − (0 + 3 + 24) = −39.
∣
2
3+2
2 −3
2
∣1 M43 =
∣
bu t
0 3
2
rD is tri
2
1
fo
M32 =
∣
∣
ot
M32 =
∣
2 3∣
1
N
∣
1
0
1
0 = (0 + 0 + 0) − (0 + 0 + 0) = 0.
−1
io n
1. 1.
0
M24 = M24 = 0.
9/9/2020
Solution 2
0
0
∣0
1
∣
A41 = (−1) ∣1 M43 =
∣
0 0
∣1
0
A43 = (−1)
1
∣ ∣
−1 ∣
4+1
2
∣
−1 ∣
2
0
0 = (0 + 0 + 0) − (0 + 0 + 0) = 0.
0
1
M41 = −M41 = 0.
−1 ∣ 1
∣ ∣
−1 ∣
4+3
0
1
0
2
0 = (0 + 0 + 0) − (0 + 0 + 0) = 0.
1
0
n
M41 =
∣
is tri bu tio
∣0
M43 = −M43 = 0.
2. 2. i. Use row 1 to compute |A|.
|A| = a11 A11 + a12 A12 + a13 A13 + a14 A14 = A11 + 0 + 2A13 − A14 . 1
0
1
−1
∣0
1
2
∣ M13 =
M14 =
∣ ∣
∣
0
1
0 −1
1
∣
2
0 0∣
1
0 1
∣ −1
M11 = M11 = 0.
∣
2
0
∣
1+1
∣
∣ −1
∣
= 0; A11 = (−1)
2
∣ ∣
0 1∣
∣ ∣
= 0; A13 = (−1)
1+3
∣
= 0; A14 = (−1)
rD
∣
∣
M13 = M13 = 0.
1+4
M14 = M14 = 0.
ot
M11 =
∣
0
∣
fo
∣0
N
Hence, |A| = 0 + 2(0) − (0) = 0 . ii. Use row 2 to compute |A|.
|A| = a21 A21 + a22 A22 + a23 A23 + a24 A24 = 2A21 + 0A22 + 0A23 + A24 . ∣0 M21 =
…
2/9
2
−1 ∣
0
1
−1
∣0
1
2
∣ ∣
∣ ∣ ∣
= 0; A21 = (−1)
2+1
M21 = 0.
9/9/2020
Solution ∣ M24 =
∣ ∣
1
0 2∣
1
0 1
∣ −1
∣ ∣
= 0; A24 = (−1)
2+4
M24 = 0.
0 1∣
Hence, |A| = 0 + 2(0) − (0) = 0 iii. Use row 3 to compute |A|. |A| = a31 A31 + a32 A32 + a33 A33 + a34 A34 = 2A31 + 0A32 + 0A33 + A34 . ∣0 M31 =
∣
M34 =
−1 ∣
0
0
1
∣0
1
2
∣
∣ M33 =
2
∣ ∣
= 0; A31 = (−1)
0 −1 ∣
2
0
1
∣ −1
0
2
∣
1
0 2∣
2
0 0
∣
∣ ∣
∣ −1
M31 = 0.
∣
1
∣
3+1
∣ ∣
∣ ∣
= 0; A33 = (−1)
3+3
M33 = 0.
∣
= 0; A34 = (−1)
3+4
M34 = 0.
0 1∣
|A| = 0 + 0 − 0 = 0.
iv. Use row 4 to compute |A|. |A| = −A41 + 0A42 + A43 + 2A44 = −A41 + A43 + 2A44 . ∣0 M41 =
M43 =
M44 =
∣
2
−1 ∣
0
∣0
1
−1 ∣
∣1
0
−1 ∣
∣
∣
1
∣
0
∣
2
0
∣1
0
−1 ∣
∣1
0
2∣
2
0
0
∣1
0
1∣
∣
∣ ∣
1
∣
∣ ∣
∣
= 0; A41 = (−1)
= 0; A43 = (−1)
= 0; A44 = (−1)
Hence, |A| = −0 + 0 + 2(0) = 0 . …
3/9
4+1
4+3
4+4
M41 = 0.
M43 = 0.
M44 = 0.
9/9/2020
Solution
i. Use row 1 to compute |B|. |B| = a11 A11 + a12 A12 + a13 A13 + a14 A14 = A11 − A12 + 2A13 + 3A14 . ∣2 M11 =
∣
0
2
3
−1
∣0
1
6
∣
A11 = (−1) ∣ M12 =
∣ ∣
2 1
∣ −3 A12 = (−1) ∣ M13 =
∣ ∣
∣ ∣
0
∣ ∣
1
2 1
∣ −3 A14 = (−1)
0
1
∣
2
0
1
3 = (36 + 0 + 3) − (−27 − 2 + 0) = 68;
−3
∣
3
∣ ∣ ∣
6
1
2
2
1
2 = (24 + 6 + 0) − (−18 + 0 + 12) = 36;
−3
0
M13 = M13 = 36.
2 0∣ 2 3
∣ ∣
0 1∣ 1+4
3 = (36 + 0 + 6) − (0 − 2 + 0) = 44;
M12 = −M12 = −68.
2 −1 0
2
∣ ∣
6
1
1+3
0
∣
3
3 −1
1+2
2
M11 = M11 = 44.
2
A13 = (−1) ∣
1+1
∣
2
∣ −3
M14 =
∣
3
2
2
1
2 = (4 − 18 + 0) − (0 + 0 + 2) = −16;
−3
0
M14 = −M14 = 16.
Hence, |B| = 44 − (−68) + 2(36) + 3(16) = 44 + 68 + 72 + 48 = 232. . ii. Use row 2 to compute |B|. |B| = a21 A21 + a22 A22 + a23 A23 + a24 A24 = 2A21 + 2A22 + 0A23 + 3A24 = 2A21 + 2A22 + 3A24 .
∣ −1 M21 =
∣ ∣ ∣
2 0
A21 = (−1)
…
4/9
2
3
3 −1 1 2+1
6
∣ ∣ ∣ ∣
−1
2
2
3 = (−18 + 0 + 6) − (0 + 1 + 24) = −37;
0
1
M21 = −(−37) = 37.
9/9/2020
Solution ∣ M22 =
∣ ∣
1 1
∣ −3 A22 = (−1) ∣ M24 =
∣ ∣
1 1
∣ −3 A24 = (−1)
2
3
3 −1 1 2+2
6
∣ ∣ ∣ ∣
2
1
3 = (18 + 6 + 3) − (−27 − 1 − 12) = 43;
−3
1
M22 = 43.
−1 2 ∣
1
−1
1
2
−3
0
∣
2
3
0
1∣
2+4
1
∣
= (2 + 9 + 0) − (−12 + 0 − 1) = 24;
M24 = 24.
Hence, |B| = 2(37) + 2(43) + 3(24) = 74 + 86 + 72 = 232. iii. Use row 3 to compute |B|. |B| = a31 A31 + a32 A32 + a33 A33 + a34 A34 = A31 + 2A32 + 3A23 − A34 . ∣ −1 M31 =
∣ ∣ ∣
2 0
A31 = (−1) ∣ −1 M32 =
∣ ∣
2
∣ −3 A32 = (−1) ∣ M33 =
∣ ∣
1 2
∣ −3 A33 = (−1) ∣ M34 =
∣ ∣
1 2
∣ −3 A34 = (−1)
2 3∣ 0 3
−1
∣ ∣
1 6∣ 3+1
0 3
0
1
2 0 = (0 − 18 + 6) − (0 + 3 + 24) = −39;
2
∣
−3
1
M32 = −M32 = −(−39) = 39.
−1 3 ∣ ∣
2
3
0
6∣
∣
1
−1
2
2
−3
0
= (12 + 9 + 0) − (−18 + 0 − 12) = 51;
M33 = 51.
−1 2 ∣ ∣
2
0
0
1∣
3+4
0 = (0 + 0 + 6) − (0 − 3 + 24) = −15;
−1
∣
1 6∣
3+3
2
M31 = −15.
2 3∣
3+2
2
∣
1
−1
2
2
−3
0
= (2 + 0 + 0) − (−12 + 0 − 2) = 16;
M34 = −M34 = −16.
|B| = −15 + 2(39) + 3(51) − (−16) = −15 + 78 + 153 + 16 = 232.
…
5/9
9/9/2020
Solution
iv. Use row 4 to compute |B|. |B| = a41 A41 + a42 A42 + a43 A43 + a44 A44 = −3A41 + 0A42 + A43 + 6A44 = −3A41 + A43 + 6A44 .
∣ −1 M41 =
∣ ∣ ∣
2
A41 = (−1)
M43 =
∣
0
3
4+1
−1
2
2
∣1
2
∣
A43 = (−1) ∣1 M44 =
3
∣
−1
∣ ∣
3 3
∣
2
0
∣1
2
3∣
4+4
2
0 = (0 + 12 + 18) − (0 − 9 − 4) = 43;
2
3
∣ ∣
1
−1
2
2
1
2
= (−2 − 3 + 12) − (6 + 6 + 2) = −7;
M43 = −M43 = −(−7) = 7.
−1 2 ∣
A44 = (−1)
∣
−1 ∣
4+3
2
= −M41 = −43.
2
∣
∣
3 −1 ∣
2
∣1
2
∣
1
−1
2
2
1
2
= (6 + 0 + 8) − (4 + 0 − 6) = 16;
M44 = 0.
Hence, |B| = −3(−43) + 7 + 6(16) = 129 + 7 + 96 = 232. . 3. 3.
We use the first row to compute |A| because it contains more zeros than the other rows. |A| = a11 A11 + a12 A12 + a13 A13 + a14 A14 = 0A11 + 0A12 + 0A13 + A14 = A14 .
∣0 M14 =
∣
1
0
0
∣1
1
∣
0 1
∣ ∣ ∣
−1 ∣
Hence, |A| = −1 . …
6/9
0
1
0
0 = (0 + 1 + 0) − (0 + 0 + 0) = 1; A14 = (−1)
1
1
1+4
M14 = −M14 = −1.
9/9/2020
Solution
4. 4.
because A is a diagonal matrix. |B| = 2(−2)(2) = −8 because B is a lower triangular matrix. |C| = 1(1)(−1) = −1 because C is an upper triangular matrix. |D| = 0 because row 2 is a zero row. |A| = 1(2)(6) = 12
|E| = 1(3)(−1)(−4)(2) = 24. |F | = 1(2)(−6)(−1) = 12. |G| = (−2)(3)(−1)(2) = 12.
Section 3.3 1. Evaluate each of the following determinants by inspection. ∣ |A| =
∣ ∣
1 0
∣ −2
2
−3 ∣
1
2
−4
6
∣ ∣ ∣
∣2 |B| =
∣ ∣
2
∣0
−2 −2 1
4∣ 4
∣ ∣
2∣
2. Use row operations to evaluate each of the following determinants.
…
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∣
1 ∣ |C| = ∣ −2
2
3
∣
1
2
∣ ∣
∣ ∣
1
−
1 2
∣ −1 ∣
9/9/2020
Solution ∣
∣
1
∣0
∣
0
∣ |D| =
∣ −2 ∣ ∣ ∣
∣
∣
0
∣
∣
4
|B| =
∣
4∣
∣
0
∣
0
0
−1
∣ −2 ∣
∣
∣
−6
0
4
0
2
−1
3
5
0
−1
2
4
−1
0
1
2
7
∣
|E| =
∣
∣
∣
−1 ∣
3
1
−2
6
10
0
∣
∣
|C| =
∣
∣ ∣
0
0∣
3
3
0
∣ 1∣
0
1
−1
0
0
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
2
−3
2
1
2
0
−2
−3
1
−2
1
−1
2
0
0
∣ ∣
2∣
1
−1
3
5
0∣
1
2
0
0
0
0
−1
2
1∣
0
0
−6
∣ |G| = ∣ −2 3 2
0
0
0
∣ ∣
7
0
6 3
N
∣5
4. Use Theorem 3.3.7
0
0
0
2
7
0
2
1
0
|B| = ∣ 0
6
3
0
0
1
3
1
2
−1
∣0
6
1
2
1
1∣ ∣
2∣ ∣
ot
|A| =
∣2
0
3 1
0
∣
∣
2∣
to find the inverse of the following matrix 0
2
−5
A = ⎜1
5
−5 ⎟
⎝
8/9
∣
∣
∣
⎛
…
∣1
0
fo
∣
0
1
0
8
⎞
⎠
6
∣ ∣
∣ ∣
1∣
3. Evaluate each of the following determinants by using its transpose. ∣1
−2 ∣
1
∣
rD
−2
∣
0
∣ −1
1
∣
−1 ∣
∣
6
∣2
2
∣
0
1
3
1
3
0
1
2
−2
0
|F | = ∣ 0 ∣
2
∣
2
∣0 ∣
−2
4∣
∣ ∣ ∣ ∣ ∣
tri bu tio n
|A| =
∣
1
is
∣2
∣ ∣ ∣ ∣. ∣ ∣ ∣
9/9/2020
…
9/9
Solution
9/9/2020
Solution
Solution
|
| | | | | | | | 1 −2 4
1 −2 4
ut
|| |
1 −2 4
() 1
R3 3 _ _
| | 1 −2
D is tri b
= 1 − 2 4 R 1 , 2_ _ − 2 1 4 R 1( − 2) + R 2_ _ − 0 5 − 4 0 2 4 0 2 4 0 2 4
4
1 −2 4
( − 2) 0 5 − 4 R 2 , 3_ _ 2 0 1 2 R 2( − 5) + R 3_ _ 0 1 2 0 5 −4
fo r
A
2 1 4
io n
1. 1. |A | = 0 because row 1 is proportional to row 3. |B | = 0 because row 1 equals row 2. |C | = 0 because row 2 is proportional to row 3. 2. 2.
N ot
2 0 1 2 = 2( − 14) = − 28. 0 0 − 14
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Solution
| | | | | | | | | | | | 0
2 3
−2 1
|B| = − 2 1 2 R 1 , 2_ _ − 3 −1 1 1 0
0
2
2
3 R (1) + R _ _ 3 1 3 −1 −1
1
1 0
1
1 0
1
1 0
1
N ot
fo r
= (1)( − 1)( − 5) = 5.
D is tri b
0 − 1 − 4 R (2) + R _ _ 0 − 1 − 4 2 3 0 2 3 0 0 −5
ut
− 0 2 3 R 1( − 3) + R 3_ _ − 0 2 3 R 2 , 3_ _ 3 −1 −1 0 −1 −4
io n
9/9/2020
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2/10
Solution
| | | | | | | | | | | | 2 −3 −2
1 2 −3 −2
2
1
0 −3 8
2
0
−2 −3 1 −2 1 −1 2
= doubt =
0
0 1 −5 −6 0 −3 5
1 2 −3 −2 0 1 −5 −6 0 −3 8
4
0 −3 5
2
2
1 2
= doubt =
−3
−2
0 −1 −5
−6
0 0
− 7 − 14
0 0 − 10 − 16
D is tri b
R 2 , 3_ _ −
4
ut
|C| =
1
io n
9/9/2020
1 2 −3 −2
= doubt = − ( − 7)(2)
0 −1 −5 −6 0 0
1
2
R 3(5) + R 4_ _
1 2 −3 −2 0 −1 −5 −6 0 0 0 0
= (14)(1)( − 1)(1)(2) = − 28.
N ot
14
fo r
0 0 −5 −8
1
2
0
2
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3/10
9/9/2020
Solution
| | | | | | | | | | | | | | 0
0 0 0
0
−2 2 0
0
1 1 0
0
= doubt = ( − 2)(6)
0 6 −6 0
0
0 0 0
0
0 1 −1 0
4 0 2 −1
0 0 0 0
R 2 , 4_ _ − 12
4 0 2 −1
1 1
− 12
0
0
0 1 −1 0
R 1( − 4) + R 2_ _
0 0 0 0
R 2 , 3_ _ 12
0
1 1 0 0 0 1 −1 0 0 0 0 0
fo r
0
4 0 2 −1 0 0 0 0
0 −4 2 −1 0 0
1 1 0 0
ut
12
1 1 0
R 1 , 2_ _
io n
4 0 2 −1
0 1 −1 0
D is tri b
|D| =
0 0 0
1 1 0
0 1 −1 0
N ot
R 2(4) + R 3_ _ 12
0
0 0 −2 −1 0 0 0
= (1)(1)( − 2)(0) = 0.
0
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4/10
9/9/2020
Solution
| |() | | | | | | | | −2 0 0 0
1 0 0 0
3 3 0 1
3 3 0 1 1 R 1 − 2 _ _ ( − 2) = doubt = 0 1 −1 0 0 1 −1 0
−2
−1 0 0 2
1 0 0 0
1 0 0 0
0 3 0 1
0 1 −1 0
0 1 −1 0
R 2 , 3_ _ 2
0 0 0 2
1 0 0 0
2
0 1 −1 0 0 0 3 1
R 2( − 3) + R 3_ _
D is tri b
0 0 0 2
0 3 0 1
io n
−1 0 0 2
ut
|E| =
= 12.
fo r
0 0 0 2
|| | | | | | | 5
0
1 −1 2
4 −1
1 −1 2
4 −1
0 −1 3
5
0
7 R __ − 0 0 1 2 1, 2 1 1 1 −2 3
7
N ot
F
0 −1 3
= 0 0 1 2 1 1 −2 3
2 − 2 6 10 0
1
2 − 2 6 10 0
1 −1 2 4 −1
() 1
R5 2
0 −1 3 5 0
−2 0 0
1 2 7
= doubt =
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5/10
9/9/2020
| | | | | | | | | | | | ()
5 2
Solution
__
1 1 −2 3 1
1 −1 2
4 −1
1 −1 2 4 −1
0 −1 3
5
0 −1 3 5 0
0
− 2 0 0 1 2 7 R 2(2) + R 4_ _ − 2 0 0 1 2 7 0 2 −4 −1 2 0 0 2 9 2 1
1
0 0 1 1 1
ut
1
1 −1 2 4
−1
0 −1 3 5
0
D is tri b
0 0
io n
1 −1 3 5 0
= doubt = − 2 0 0 1 2 7 R 4 , 5_ _ 0 0 0 5 − 12
−1
0 −1 3 5
0
N ot
1 −1 2 4
fo r
0 0 0 −1 −6
1 −1 2 4
−1
0 −1 3 5
0
2 0 0 1 2 7 R 4(5) + R 5_ _ 2 0 0 1 2 7 0 0 0 −1 −6 0 0 0 −1 −6 0 0 0 5 − 12
0 0 0 0 − 42
= 2( − 1)(1)( − 1)( − 42) = − 84.
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6/10
9/9/2020
Solution
|| | | | | 1 2 0 5
AT
=
0 7 6 3 0 0 3 1
R 1( − 1) + R 4_ _
1 2 0 2
| | 0 7 6 3 0 0 3 1
0 0 0 −3
||| | | | | | | | | |
|B| =
BT
1 2 0 1 0
1 2 0 1 0
0 7 6 3 6
0 1 0 −1 1
= 0 0 3 1 1 R 2 , 5_ _ − 0 0 3 1 1 0 2 0 2 2 0 2 0 2 2 0 1 0 −1 1
1 2 0 1
0 7 6 3 6
0
1 R 3( − 2) + R 5_ _ 0
0 0 6 10 − 1
1 2 0 1
0
0 1 0 −1 1
− 0 0 3 1 0 0 0 4
N ot
= doubt = 0 0 3 1 0 0 0 4
fo r
0 1 0 −1 1
1 2 0 1
0
0 1 0 −1 1
1 R 4( − 2) + R 5_ _ − 0 0 3 1 0 0 0 0 4
0 0 0 8 −3
4. 4.
= − 63.
io n
=
ut
A
D is tri b
3. 3.
1 2 0 5
1
= 36
0
0 0 0 0 −3
By (3.1.9), we have
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7/10
9/9/2020
Solution
| | | | | | 0 8
A 21 = −
A 31 =
| | | | | |
| | | | | |
1 −5 1 5 = 40, A 12 = − = − 13, A 13 = = − 5, 1 8 1 0
2 −5 0 8
0 −5 0 2 = − 16, A 22 = = 5, A 23 = − = 2, 1 8 1 0
2 −5
0 −5 0 2 = 15, A 32 = − = − 5, A 33 = = − 2. 5 −5 1 −5 1 5
io n
5 −5
ut
A 11 =
adj(A) =
(
A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33
D is tri b
This, together with (3.3.11), implies
)( =
40 − 16 15
− 13
5
−5
2
)
−5 . −2
adj(A) = − adj(A) =
N ot
A −1 =
1
fo r
By computation, |A | = − 1. It follows from Theorem 3.3.7 that
|A|
(
− 40 16 − 15 13 − 5
5
5
2
−2
)
.
Section 3.4 1. For each of the following matrices, use its determinant to determine whether it is invertible.
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E =
0
0
2
1
2
0
−2 −3 1 −2
(
(
1
−1 0
2
1
4
1
−2
4
−2 −1 −4
0
)
B=
) ( D=
1
−1
3
5
0
0
0
0
0
0
−2
0
−1
2
1
3
0
0
2
0
0
−6 6 0
1
)
(
0 −1
3
5
2
1 −1
2
4
−1
0
0
1
2
7
1
1
−2
3
1
6
10
4
0 −2
0
1
3
−2
1
2
1
−1 −1
)
)
ut
C =
(
0
F=
(
1
0
0
0
−2 2
0
0
0
6 −6
4
0
2
fo r
A =
1
io n
Solution
D is tri b
9/9/2020
0
−1
)
N ot
2. For each of the following matrices, find all real numbers x such that it is invertible.
A=
(
x
0
0
0
2
x
2
0
−2 −3 1 −2 1
0
0
1
)
( ) 2
B=
1
x
4
−2 4
−2 −1 x
3. Let A be a 4 × 4 matrix. Assume that |A | = 2. Compute each of the following determinants. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
9/10
9/9/2020
Solution
1. | − 3A|
| | 3. |A | 4. |A | 2. A − 1 T 3
6. (2( − A) T)
−1
io n
| |
ut
T
( ) ( ) 2 1 3
4. Compute |A | + | B| if A =
D is tri b
| |
5. (2A − 1)
1 0 3
and B =
1 4 7
2 1
3
1 0
3
0 1 −1
( ) ( ) 0 0 2
1
0 0 2
−1 0 1 2
3
0 0 1
0
1 4 3
and B =
N ot
1
A=
fo r
5. Compute |A | + | B| if
.
0
1 4 3
−2 2 0 1
.
−2 2 0 1
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10/10
9/9/2020
Solution
Solution
|
| | | | | | | | 1 −2 4
1 −2 4
ut
|| |
1 −2 4
() 1
R3 3 _ _
| | 1 −2
D is tri b
= 1 − 2 4 R 1 , 2_ _ − 2 1 4 R 1( − 2) + R 2_ _ − 0 5 − 4 0 2 4 0 2 4 0 2 4
4
1 −2 4
( − 2) 0 5 − 4 R 2 , 3_ _ 2 0 1 2 R 2( − 5) + R 3_ _ 0 1 2 0 5 −4
fo r
A
2 1 4
io n
1. 1. |A | = 0 because row 1 is proportional to row 3. |B | = 0 because row 1 equals row 2. |C | = 0 because row 2 is proportional to row 3. 2. 2.
N ot
2 0 1 2 = 2( − 14) = − 28. 0 0 − 14
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1/10
Solution
| | | | | | | | | | | | 0
2 3
−2 1
|B| = − 2 1 2 R 1 , 2_ _ − 3 −1 1 1 0
0
2
2
3 R (1) + R _ _ 3 1 3 −1 −1
1
1 0
1
1 0
1
1 0
1
N ot
fo r
= (1)( − 1)( − 5) = 5.
D is tri b
0 − 1 − 4 R (2) + R _ _ 0 − 1 − 4 2 3 0 2 3 0 0 −5
ut
− 0 2 3 R 1( − 3) + R 3_ _ − 0 2 3 R 2 , 3_ _ 3 −1 −1 0 −1 −4
io n
9/9/2020
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2/10
Solution
| | | | | | | | | | | | 2 −3 −2
1 2 −3 −2
2
1
0 −3 8
2
0
−2 −3 1 −2 1 −1 2
= doubt =
0
0 1 −5 −6 0 −3 5
1 2 −3 −2 0 1 −5 −6 0 −3 8
4
0 −3 5
2
2
1 2
= doubt =
−3
−2
0 −1 −5
−6
0 0
− 7 − 14
0 0 − 10 − 16
D is tri b
R 2 , 3_ _ −
4
ut
|C| =
1
io n
9/9/2020
1 2 −3 −2
= doubt = − ( − 7)(2)
0 −1 −5 −6 0 0
1
2
R 3(5) + R 4_ _
1 2 −3 −2 0 −1 −5 −6 0 0 0 0
= (14)(1)( − 1)(1)(2) = − 28.
N ot
14
fo r
0 0 −5 −8
1
2
0
2
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9/9/2020
Solution
| | | | | | | | | | | | | | 0
0 0 0
0
−2 2 0
0
1 1 0
0
= doubt = ( − 2)(6)
0 6 −6 0
0
0 0 0
0
0 1 −1 0
4 0 2 −1
0 0 0 0
R 2 , 4_ _ − 12
4 0 2 −1
1 1
− 12
0
0
0 1 −1 0
R 1( − 4) + R 2_ _
0 0 0 0
R 2 , 3_ _ 12
0
1 1 0 0 0 1 −1 0 0 0 0 0
fo r
0
4 0 2 −1 0 0 0 0
0 −4 2 −1 0 0
1 1 0 0
ut
12
1 1 0
R 1 , 2_ _
io n
4 0 2 −1
0 1 −1 0
D is tri b
|D| =
0 0 0
1 1 0
0 1 −1 0
N ot
R 2(4) + R 3_ _ 12
0
0 0 −2 −1 0 0 0
= (1)(1)( − 2)(0) = 0.
0
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4/10
9/9/2020
Solution
| |() | | | | | | | | −2 0 0 0
1 0 0 0
3 3 0 1
3 3 0 1 1 R 1 − 2 _ _ ( − 2) = doubt = 0 1 −1 0 0 1 −1 0
−2
−1 0 0 2
1 0 0 0
1 0 0 0
0 3 0 1
0 1 −1 0
0 1 −1 0
R 2 , 3_ _ 2
0 0 0 2
1 0 0 0
2
0 1 −1 0 0 0 3 1
R 2( − 3) + R 3_ _
D is tri b
0 0 0 2
0 3 0 1
io n
−1 0 0 2
ut
|E| =
= 12.
fo r
0 0 0 2
|| | | | | | | 5
0
1 −1 2
4 −1
1 −1 2
4 −1
0 −1 3
5
0
7 R __ − 0 0 1 2 1, 2 1 1 1 −2 3
7
N ot
F
0 −1 3
= 0 0 1 2 1 1 −2 3
2 − 2 6 10 0
1
2 − 2 6 10 0
1 −1 2 4 −1
() 1
R5 2
0 −1 3 5 0
−2 0 0
1 2 7
= doubt =
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5/10
9/9/2020
| | | | | | | | | | | | ()
5 2
Solution
__
1 1 −2 3 1
1 −1 2
4 −1
1 −1 2 4 −1
0 −1 3
5
0 −1 3 5 0
0
− 2 0 0 1 2 7 R 2(2) + R 4_ _ − 2 0 0 1 2 7 0 2 −4 −1 2 0 0 2 9 2 1
1
0 0 1 1 1
ut
1
1 −1 2 4
−1
0 −1 3 5
0
D is tri b
0 0
io n
1 −1 3 5 0
= doubt = − 2 0 0 1 2 7 R 4 , 5_ _ 0 0 0 5 − 12
−1
0 −1 3 5
0
N ot
1 −1 2 4
fo r
0 0 0 −1 −6
1 −1 2 4
−1
0 −1 3 5
0
2 0 0 1 2 7 R 4(5) + R 5_ _ 2 0 0 1 2 7 0 0 0 −1 −6 0 0 0 −1 −6 0 0 0 5 − 12
0 0 0 0 − 42
= 2( − 1)(1)( − 1)( − 42) = − 84.
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6/10
9/9/2020
Solution
|| | | | | 1 2 0 5
AT
=
0 7 6 3 0 0 3 1
R 1( − 1) + R 4_ _
1 2 0 2
| | 0 7 6 3 0 0 3 1
0 0 0 −3
||| | | | | | | | | |
|B| =
BT
1 2 0 1 0
1 2 0 1 0
0 7 6 3 6
0 1 0 −1 1
= 0 0 3 1 1 R 2 , 5_ _ − 0 0 3 1 1 0 2 0 2 2 0 2 0 2 2 0 1 0 −1 1
1 2 0 1
0 7 6 3 6
0
1 R 3( − 2) + R 5_ _ 0
0 0 6 10 − 1
1 2 0 1
0
0 1 0 −1 1
− 0 0 3 1 0 0 0 4
N ot
= doubt = 0 0 3 1 0 0 0 4
fo r
0 1 0 −1 1
1 2 0 1
0
0 1 0 −1 1
1 R 4( − 2) + R 5_ _ − 0 0 3 1 0 0 0 0 4
0 0 0 8 −3
4. 4.
= − 63.
io n
=
ut
A
D is tri b
3. 3.
1 2 0 5
1
= 36
0
0 0 0 0 −3
By (3.1.9), we have
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7/10
9/9/2020
Solution
| | | | | | 0 8
A 21 = −
A 31 =
| | | | | |
| | | | | |
1 −5 1 5 = 40, A 12 = − = − 13, A 13 = = − 5, 1 8 1 0
2 −5 0 8
0 −5 0 2 = − 16, A 22 = = 5, A 23 = − = 2, 1 8 1 0
2 −5
0 −5 0 2 = 15, A 32 = − = − 5, A 33 = = − 2. 5 −5 1 −5 1 5
io n
5 −5
ut
A 11 =
adj(A) =
(
A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33
D is tri b
This, together with (3.3.11), implies
)( =
40 − 16 15
− 13
5
−5
2
)
−5 . −2
adj(A) = − adj(A) =
N ot
A −1 =
1
fo r
By computation, |A | = − 1. It follows from Theorem 3.3.7 that
|A|
(
− 40 16 − 15 13 − 5
5
5
2
−2
)
.
Section 3.4 1. For each of the following matrices, use its determinant to determine whether it is invertible.
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8/10
E =
0
0
2
1
2
0
−2 −3 1 −2
(
(
1
−1 0
2
1
4
1
−2
4
−2 −1 −4
0
)
B=
) ( D=
1
−1
3
5
0
0
0
0
0
0
−2
0
−1
2
1
3
0
0
2
0
0
−6 6 0
1
)
(
0 −1
3
5
2
1 −1
2
4
−1
0
0
1
2
7
1
1
−2
3
1
6
10
4
0 −2
0
1
3
−2
1
2
1
−1 −1
)
)
ut
C =
(
0
F=
(
1
0
0
0
−2 2
0
0
0
6 −6
4
0
2
fo r
A =
1
io n
Solution
D is tri b
9/9/2020
0
−1
)
N ot
2. For each of the following matrices, find all real numbers x such that it is invertible.
A=
(
x
0
0
0
2
x
2
0
−2 −3 1 −2 1
0
0
1
)
( ) 2
B=
1
x
4
−2 4
−2 −1 x
3. Let A be a 4 × 4 matrix. Assume that |A | = 2. Compute each of the following determinants. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
9/10
9/9/2020
Solution
1. | − 3A|
| | 3. |A | 4. |A | 2. A − 1 T 3
6. (2( − A) T)
−1
io n
| |
ut
T
( ) ( ) 2 1 3
4. Compute |A | + | B| if A =
D is tri b
| |
5. (2A − 1)
1 0 3
and B =
1 4 7
2 1
3
1 0
3
0 1 −1
( ) ( ) 0 0 2
1
0 0 2
−1 0 1 2
3
0 0 1
0
1 4 3
and B =
N ot
1
A=
fo r
5. Compute |A | + | B| if
.
0
1 4 3
−2 2 0 1
.
−2 2 0 1
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10/10
9/9/2020
Solution
Solution 1. 1.
∣ ∣ ∣ A = ∣ ∣ ∣ ∣
∣
∣
∣
∣
∣
∣1 ∣
∣0 ∣ ∣
0
1 0
0
2
1
0 2
−2
−3 1
1
−1 0
−2
1
2
0
∣
∣0
1/6
1
0
R2 (−2) + R3 ∣ ––––––––––––– ––––––––––––– ∣
0
−2
0
∣
1
0
1
2
0
−2
0
= (1)(1)(1)(4) = 4 ≠ 0,
∣
ot
∣
N …
∣
R4 (3) + R3 ∣ – –––––––––– –––––––––– – 0 ∣ ∣
∣ −3 −1 ∣
4
∣
∣0
∣
1
0
2
∣ −3 −1 ∣
∣
0
∣ −3 −1 ∣
1
1
∣0
1
−2
−2
∣
−2
2
2
2
∣
1
∣
2
∣
∣1
1
∣
∣1
0
0
∣1
∣
∣
−2
0
∣
∣
∣0 ∣ T∣ = A = ∣ ∣ ∣ ∣ −2 0 ∣ ∣ ∣ ∣
−3 −1 ∣ 7
∣
∣
∣
0
∣
0
∣
∣0
∣
∣0
0
∣
∣0
∣
so A is invertible.
2
1
io n
∣
∣
bu t
∣
∣
R3 (2) + R4 – –––––––––– –––––––––– –
rD is tri
∣
∣
fo
∣
∣ ∣ ∣
Solution ∣0
∣
∣
∣
∣
∣
1
−1
∣
∣
∣
∣
∣
∣
∣
∣0
5
2
−1
2
4
−1
0
1
2
7
1
−2
3
1
−2
6
10
4
∣B∣ = ∣ 0 ∣
3
1
∣1
∣
0
−1 2 −1 3
−∣0 ∣ ∣
0
∣0
4 5
0
1
2
0
2
−4
−1
∣0
0
2
9
− ∣0 ∣
−1 ∣ 2
2
7
0
2
9
6
9
6
4
−1
3
5
2
0
1
2
7
1
−2
3
1
−2
6
10
4
2
∣ ∣ ∣ ∣ ∣
∣ ∣
7 2 6
∣ R2 (2) + R4 –––––––––– –––––––––– – ∣– ∣ ∣
∣
∣
∣
−1 ∣
∣1
∣
∣
0
−1 ∣
−1 2
4
−1 3
5
2
∣ R3 (−1) + R4 − ∣ 0 –––––––––––– –––––––––––– – ∣– ∣ 0 ∣ ∣
0
1
2
7
0
2
9
6
∣
0
0
0
0
∣0
2
∣ ∣ ∣ C = ∣ ∣ ∣
1
∣
ot
∣
∣
∣
1
∣ −2
4
∣
−2
4
∣
∣2
∣ R1 (1) + R3 − 1 ∣ ––––––––––– ∣ ––––––––––– ∣ ∣0 −1 −4
N
∣
2/6
2
∣ ∣
∣ = 0, ∣ ∣ ∣
fo
so B is not invertible. C is not invertible because
…
∣0
∣
5
0
0
−1 ∣
−1
∣ R1, 2 − ∣ 0 –– –– –– – –– ∣– ∣ 1 ∣ ∣
3
1
2
∣
−1
0
0
∣
4
∣
∣
∣
2
∣
∣1
∣1
∣
−1
∣
R2 (−1) + R4 –––––––––––– –––––––––––– – –
∣
is tri bu tio
∣
∣
rD
∣
n
9/9/2020
1
4∣
−2 4 0
∣ ∣
0∣
= 0.
Solution ∣
∣
∣
0
1
∣ ∣ ∣ D = −2 ∣ ∣ ∣ ∣
∣
∣ ∣1
−
∣ ∣
0
3
∣
∣
∣
∣
R1, 3 ∣ ––––– ––––– −1 −1 ∣ 1
1
2
−
∣ ∣
−2
1
2
0
1
3
−1 −1 ∣ −1
0
1
3
∣0
−1 −1 ∣
1
∣1
∣
R2 (1) + R3 ∣ ––––––––––– ––––––––––– ∣
−
∣ ∣
0
∣0
−1
0
0
3
∣ ∣ ∣ ∣
2
0 x
0 0 2 0
−2 −3 1 −2 1
0
0 1
∣ x 2 −2 1 ∣
∣
∣
∣ ∣ ∣
T
= A
∣ 0 x −3 0 ∣ ∣ ∣ ∣
∣
= 3 ≠ 0,
∣
bu t
∣ ∣
∣
is tri
|A| =
x
0 2 1
rD
∣
R1 (2) + R2 ∣ ––––––––––– ––––––––––– ∣
−1 −1 ∣
so D is invertible. Because it contains a zero row, |E| = 0 and E is not invertible. |F | = 1(2)(−6)(−1) = 12 ≠ 0, and F is invertible. 2. 2. ∣
∣
io n
9/9/2020
0
∣ ∣
∣ 0 0 −2 1 ∣
∣ x −3 0 ∣
= x
∣ ∣
2
1
0
∣ ∣
∣ 0 −2 1 ∣
fo
= x[x − (−6)] = x(x + 6).
N
ot
Let |A| = 0. Then x = 0 or x = −6. Hence, when x ≠ 0 and x ≠ −6, |A| ≠ 0 and A is invertible.
…
3/6
9/9/2020
Solution
|B| =
∣ ∣
2
x
1
∣1 −
∣ ∣
0
∣
∣
∣
4
−2
−2 x
4∣ 4
∣ ∣
−1 x ∣
∣x+4 = −(1) ∣ ∣ ∣ −5 x+8 ∣ ∣
−4
−5
= doubt =
∣
4
x+4
∣0
1
R1, 2 − 2 ∣ ––––– ∣ ––––– ∣ −2 −1 x ∣ −2
∣ −2
4∣
∣ ∣ x+8∣ −4
2
= −[(x + 4)(x + 8) − (−4)(−5)] = −(x 2
– 2 2 − 24] = (2√6) − (x + 6)
– – = [2√6 − (x + 6)][2√6 + (x + 6)]. –
+ 12x + 12)
bu t
= −[ (x + 6)
io n
∣
–
is tri
Let |B| = 0. Then [2√6 − (x + 6)][2√6 + (x + 6)] = 0. This implies – x = −6 + 2√6 –
–
– or x = −6 − 2√6.
3. 3. 4
∣−3A∣ = (−3) ∣A∣ = 162. ∣ ∣ ∣ ∣
2.
∣ −1 ∣ A = ∣ ∣
3.
∣ ∣A
4.
∣ ∣A ∣ ∣ = ∣ ∣A|
5.
T 4∣ ∣ −1 −1 ∣ −1 ∣ ∣ ∣ (2A ) = 2A = 2 A = ∣ ∣ ∣ ∣ ∣ ∣
1 2
∣ ∣ = ∣ ∣A∣ ∣ = 2. 3
= 8.
.
ot
3
=
N
−1
1 |A|
fo
1.
16
|A|
= 8.
−1
6.
T ∣ (2(−A) ) ∣
∣ ∣
=
=
4. 4. …
4/6
rD
Hence when x ≠ −6 − 2√6 and x ≠ −6 + 2√6, ∣∣B∣∣ ≠ 0 and B is invertible.
1 T
∣2(−A) ∣
∣ ∣
=
1 4
16(−1) ∣A∣ ∣ ∣
1 T
4 2 ∣ ∣(−A)
=
1 32
.
∣ ∣
=
1 16|−A|
Solution
|A| + |B| =
∣ ∣
1
∣1 ∣ =
∣ ∣
1
3∣
0
3
4
7∣
=
∣
3
1
0
3
∣0
1
∣
1
0
3
0
3
∣1
5
∣
∣ ∣ ∣
∣1
∣
∣ R1, 2 − 2 ∣ –– –– –– – – –– ∣1 6∣ ∣
∣1
0
3∣
1
3
5
6∣ 3
0
1
−3
∣0
5
3
∣
∣
∣
0
∣
∣
∣ ∣ ∣
is t
= doubt = −
∣
7−1∣
4+1
1
∣
−1 ∣ 3
3∣
∣
+
1
∣
1
1
∣
∣2
2
∣1+0 ∣2
∣
n
∣2
rib ut io
9/9/2020
∣1
N
ot
fo
5. 5.
…
5/6
3
1
−3
rD
∣ R2 (−5) + R3 − 0 ∣ –––––––––––– –––––––––––– – – ∣0
0
0
∣ ∣ ∣
18 ∣
= −18.
Solution ∣
∣
∣
∣
∣
∣
∣
2∣
∣
∣
∣
0 1
2∣
1 4
3
∣
2 0
1∣
∣
∣
∣
∣ −2
1
0
∣
∣ −1 + 3 ∣ ∣ ∣ ∣1 ∣
=
∣
∣ 1∣
0
1 4
3
2 0
1∣
∣ −2
0
2
0+0
1+0
0
1
4
3
−2
2
0
1
2
0 0
∣
∣
∣
∣
2 + 1∣
0 −2 ∣
=
∣
0
1
4
3
∣2
3
3
1
∣
∣0
2
0 −2 ∣
0
1
2
∣ ∣
∣ R2, 3 −
1
4
0
∣0
0
7
5
∣1
2
0 −2 ∣
1
4
0
0
0
1
2
∣0
0
0 −9 ∣
∣
∣
∣0 ∣ ∣
0
∣0
∣ ∣ ∣
= 9.
∣
0 1
∣ 3∣
0
1 4
3
2
0
0
1
1
4
−1 3
2
0 −2 ∣
1
4
0
0
1
2
0
7
5
ot
∣
∣
–– –– –– – – ––
N
∣0
∣0
∣
0
∣
∣0
∣1
2
∣ −2
R1 (−2) + R4 ∣ ––––––––––––– ∣ ––––––––––––– 0 ∣ ∣
∣
2∣
∣
∣
∣
∣
∣
0 0
∣
∣1
∣
∣
∣
1
∣
2
∣
−
∣
3
1
∣1
∣
∣
2∣
0
∣
∣
+
0 0
∣ ∣
2 0
1∣
−2 ∣ 2
∣ ∣
3
R3 (1) + R4 ∣ ––––––––––– ––––––––––– ∣
5
∣
∣
rD
∣0
∣
∣
1
fo
=
6/6
0 0
∣ ∣ ∣ ∣ ∣ −1 A + B = ∣ ∣ ∣ ∣ ∣ 0 ∣ ∣ ∣ ∣ ∣
∣
…
1
∣ ∣ ∣ ∣
n
∣
∣
rib ut io
∣
is t
9/9/2020
R3 (−7) + R4 –––––––––––– –––––––––––– – –
9/9/2020
A.4 Systems of linear equations
A.4 Systems of linear equations Section 4.1
2 3
bu t
– − −√2x + 6
y = 4 − 3z
3x1 + 2x2 + 4x3 + 5x4 = 1 2xy + 3yz + 5z = 8
rD is tri
c. d. e.
io n
1. Determine which of the following equations are linear. a. x + 2y + z = 6 b. 2x − 6y + z = 1
2. For each of the following systems of linear equations, find its coefficient and augmented matrices 1.
{
2x1 − x2 = 6 4x1 + x2 = 3
⎧ ⎪ −x1 + 2x2 + 3x3 = 4 ⎨ 3x1 + 2x2 − 3x3 = 5 ⎩ ⎪ 2x1 + 3x2 − x3 = 1
fo
2.
⎨ 2x2 − 5x3 − 8 = 0 ⎩ ⎪ 3x1 + 2x2 − x3 = 4
N
3.
ot
⎧ x1 − 3x3 − 4 = 0 ⎪
3. For each of the following augmented matrices, find the corresponding linear system. ⎛
1
⎜ −1 ⎝
…
1/3
0
2 1 −1
3∣ 0
∣ ∣
1∣
1
⎞
−2 ⎟ 0
⎠
⎛
2
⎜0 ⎝
0
1 −1 3
∣ ∣ ∣
1
⎞
−2 ⎟
∣ −1
⎠
9/9/2020
A.4 Systems of linear equations →
→
4. For each of the following linear systems, write it into the form A X combination of the column vectors of A.
=
b
and express
→ b
as a linear
⎧ ⎪ −x1 + x2 + 2x3 = 3
a.
⎨ 2x1 + 6x2 − 5x3 = 2 ⎩ ⎪ −3x1 + 7x2 − 5x3 = −1
c.
⎨ −2x1 + 4x2 − 3x3 = 0 ⎩ ⎪ 3x1 + 6x2 − 8x3 = 0, {
is tri bu tio
b.
n
⎧ ⎪ x1 − x2 + 4x3 = 0
x1 − 2x2 + x3 = 0 −x1 + 3x2 − 4x3 = 0
5. For each of the following points, P1 (6, −
1 2
) , P 2 (8, 1), P 3 (2, 3),
is a solution of the system
x − 2y = 7,
rD
{
and P4 (0, 0), verify whether it
x + 2y = 5.
fo
6. Consider the system of linear equations
ot
⎧ ⎪ −x1 + x2 + 2x3 = 3
N
⎨ 2x1 + 6x2 − 5x3 = 2 ⎩ ⎪ −3x1 + 7x2 − 5x3 = −1
1. Verify that (1, 3, − 1) is a solution of the system. 2. Verify that ( 56 , 76 , 43 ) is a solution of the system. →
3. Determine if the vector b = (3, 2, − 1) coefficient matrix of the system.
…
2/3
T
is a linear combination of the column vectors of the
9/9/2020
A.4 Systems of linear equations →
T
4. Determine if the vector b = (3, 2, − 1) belongs to the spanning space of the column vectors of the coefficient matrix of the system. 7. Solve each of the following systems. 1.
{
x1 − x2 = 6
2.
{
io n
x1 + x2 = 5 x1 + 2x2 = 1 2x1 + x2 = 2 {
x1 + 2x2 = 1
bu t
3.
N
ot
fo
rD
is tri
x1 + 2x2 = 2
…
3/3
9/9/2020
Solution
Solution a), b), c), d) are linear.
( )| ( ) ( )| ( ) −1 2 3
2. A =
3. A =
)
−1 2 3 4
3 2 − 3 , (A → b) = 2 3 −1
3 2 −3 5 2 3 −1 1
1 0 −3
1 0 −3 4
0 2 − 5 , (A → b) = 3 2 −1
0 2 −5 8 3 2 −1 4
ot
x 1 + 2x 2 + 3x 3 = 1 − x1 + x2 = − 2
N
{
fo
3. 3.
4. 4.
…
1/6
n
4 1 3
tio
2 −1 6
bu
4 1
, (A B) =
tri
( ) | ( 2 −1
is
1. A =
rD
1. 1. 2. 2.
− x2 + x3 = 0
{
2x + y = 1 −y = − 2 3y = − 1
9/9/2020
Solution
( ) () ( ) () −1 1 2
→ 2 6 −5 , X = −3 7 −5
() () 2
−1
x3
( ) () −1 2
−3
→ , c2 =
0
−1 3 −4
()
fo
)
x1
→
, X=
x 2 , and 0 = x3
() 0
→
N
(
1 −2 1
→ −2 , c = 2 3
rD
4
→ → → − 3 . Then x c + x c + x c = → b. 1 1 2 2 3 3 −8
() ()
tri
x3
|
1
is
x 2 , and → b=
→ → → 0 . Then A X = b. Let c 1 = 0
0 . Then the system can be rewritten as 0 →
→
AX = 0.
…
2/6
6 , 7
tio
()
bu
() x1
→ −2 4 −3 , X = 3 6 −8
→ and c 3 =
→ Let a 1 =
1
n
→ → → − 5 . Then x c + x c + x c = → b. 1 1 2 2 3 3 −5
1 −1 4
c. Let A =
x 2 , and → b=
→ → → . Then AX = b. Let c 1 =
2
→ and c 3 =
b. Let A =
3
ot
a. Let A =
x1
( ) ( )
( )
→ → −2 1 , a2 = , and a 3 = . Then the system can be rewritten as −1 3 −4 1
−1 4 6
,
9/9/2020
Solution
→ → → → x 1a 1 + x 2a 2 + x 3a 3 = 0.
5. 5.
( ) 1
Because 6 − 2 − 2
( ) ( ) 1
= 7 and 6 + 2 − 2
= 5,
6, −
1
2
is a solution. Because
8 − 2(2) = 6 ≠ 7, (8, 1) is not a solution. Because 2 − 2(3) = 2 − 6 = − 4 ≠ 7, (2, 3) is not a solution. Because 0 − 2(0) = 0 ≠ 7, (0, 0) is not a solution. 6. 6. 1. Because − 1 + 3 + 2( − 1) = 0 ≠ 3, (1, 3, − 1) is not a solution. 2. Because
{ (
5
, 6
7
, 6
4 3
→ 3. Let c 1 =
)
() () () () () () () 5
−6 +
2
−3
5
7
6
+6
6
5
6
4
+2
3
7
− 5 + 7 + 16
−5
6
+7
=
7
6
−5
6
4
=
3
4
3
5 3
18
=
+7−
= −
= 3,
6
15 6
20
= 7 − 5 = 2,
3
+
49 6
−
40 6
= − 1,
is a solution of the system.
( ) () −1 2
→ , c2 =
−3
→
1
→ 6 , and c 3 = 7
() 2
− 5 . Because −5
(
5
, 6
7
, 6
4 3
)
is a solution of the system, by
Theorem 4.1.1, b = (3, 2, − 1) T is a linear combination of c 1, c 2, c 3.
…
3/6
9/9/2020
Solution
→
4. Because b =
(
5 6
,
7 6
,
4 3
)
{
7. 7. 1. Adding the two equations implies 2x 1 = 11 and x 1 = x2 = x1 − 6 =
11 2
1
− 6 = − 2 , so (x 1, x 2) =
(
11 2
, −
1 2
)
11 2
. By the first equation, we have
is a solution.
2. The first equation minus 2 times the second equation implies (x 1 + 2x 2) − 2(2x 1 + x 2) = 1 − 2(2) and x 1 = 1. Substituting x 1 = 1 into the second equation, x 2 = 2 − 2x 1 = 2 − 2(1) = 0. Hence (x 1, x 2) = (1, 0) is a solution. 3. This system has no solutions.
Section 4.2 1. Consider each of the following systems.
i.
ii.
…
4/6
{ {
x 1 + x 2 − 3x 3 = 2 − x 2 − 6x 3 = 4 − x3 = 1 x 1 − 2x 2 + 3x 3 + x 4 = − 1 x2 − x3 + x4 = 4 − x4 = 2
}
→ → → is a solution of the system, by Theorem 4.1.1, b ∈ span c 1 , c 2 , c 3 . →
9/9/2020
Solution
iii.
iv.
{ {
− x 1 − 2x 2 − 2x 3 + x 4 − 3x 5 = 4 − 2x 3 + 3x 4 = 1 − 4x 4 + 2x 5 = 5 − 3x 1 − 2x 2 − 2x 3 = 2 x2 − x3 = 2 0x 3 = 4
a. Find the coefficient and augmented matrices of the system. b. Find the basic variables and free variables of the system. c. Solve the system by using back-substitution if it is consistent, and express its solution as a linear combination. →
d. Determine if the vector b on the right side of the system is a linear combination of the column vectors of the coefficient matrix of the system. 2. Solve each of the following systems by using Gaussian elimination and express the solutions as linear combinations if they have infinitely many solutions.
a.
…
5/6
{
x 1 − 2x 2 + 3x 3 = 6 2x 1 + x 2 + 4x 3 = 5 − 3x 1 + x 2 − 2x 3 = 3
9/9/2020
Solution
b.
c.
d.
e.
f.
…
6/6
{ { { { {
x 1 + x 2 − 2x 3 = 3 2x 1 + 3x 2 − x 3 = − 4 − 2x 1 − 3x 2 − x 3 = 4 x 1 + 2x 2 + 3x 3 = 4 4x 1 + 7x 2 + 6x 3 = 17 2x 1 + 5x 2 + 12x 3 = 7 x 1 + 2x 2 − x 3 + 3x 4 = 5 3x 1 − x 2 + 4x 3 + 2x 4 = 1 − x 1 + 5x 2 − 6x 3 + 4x 4 = 9 x 1 − 2x 2 = 3 2x 1 − x 2 = 0 − x 1 + 4x 2 = − 1 − x 1 + 2x 2 − 3x 3 = − 4 2x 1 − x 2 + 2x 3 = 2 x1 + x2 − x3 = 2
9/9/2020
Solution
Solution 1. 1. a. The coefficient and the augmented matrices are
( )| (
)
1 1 −3
1 1 −3 2
0 − 1 − 6 , (A → b) = 0 0 −1
0 −1 −6 4 . 0 0 −1 1
D is tri b
A=
ut io n
i.
b. Because A is a row echelon matrix, the basic variables are x 1, x 2, and x 3. There are no free variables. c. By the third equation, x 3 = − 1. By the second equation,
and by the first equation,
fo r
x 2 = − 6x 3 − 4 = − 6( − 1) − 4 = 2
N ot
x 1 = 2 − x 2 + 3x 3 = 2 − 2 + 3( − 1) = − 3.
Hence, (x 1, x 2, x 3) = ( − 3, 2, − 1) is a solution. → d. Because the system has a solution ( − 3, 2, − 1), b = (2, 4, 1) T is a linear combination of the column vectors of A. ii. Processing math: 100%
a. The coefficient and the augmented matrices are
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1/15
9/9/2020
Solution
A=
(
1 −2 3
1
0 1 −1 1 0 0
0 −1
)| ( →
, (A b) =
1 −2 3
1 −1
0 1 −1 1 0 0
4
0 −1 2
)
.
ut io n
b. x 1, x 2, x 4 are basic variables and x 3 is a free variable. c. By the third equation, x 4 = − 2. Let x 3 = t. Then by the second equation, x 2 = 4 − x 4 + x 3 = 4 + 2 + x 3 = 6 + t. By the first equation,
D is tri b
x 1 = − 1 + 2x 2 − 3x 3 − x 4 = − 1 + 2(6 + t) − 3t − ( − 2) = 13 − t.
()()()()() x1 x2 x3
13 6
=
0
−t
+
−2
t
=
t
0
6 0
−1
+t
−2
fo r
x4
13
1 1
.
0
→
N ot
d. Because the system has infinitely many solutions, b = ( − 1, 4, 2) T is a linear combination of the column vectors of A. iii.
a. The coefficient and the augmented matrices are
A= Processing math: 100%
(
−1 −2 −2 1 −3 0
0 −2 3
0
0
0
0 −4 2
)| ( →
, (A b) =
−1 −2 −2 1 −3 4 0
0 −2 3
0
0
)
0 1 . 0 −4 2 5
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2/15
9/9/2020
Solution
b. x 1, x 2, x 4 are basic variables and x 2 and x 5 are free variables. c. Let x 2 = s and x 5 = t. By the third equation, 5
1
− 4x 4 = 5 − 2x 5 and x 4 = − 4 + 2 t.
19
3
)
5 1 15 3 19 3 + t =1+ − t= − t 4 2 4 2 4 2
and x 3 = − 8 + 4 t. By the first equation,
D is tri b
(
− 2x 3 = 1 − 3x 4 = 1 − 3 −
ut io n
By the second equation, we have
x 1 = − 4 − 2x 2 − 2x 3 + x 4 − 3x 5
(
19
3
) (
5
1
)
= − 4 − 2s − 2 − 8 + 4 t + − 4 + 2 t − 3t 19
3
5
1
1
N ot
fo r
= − 4 − 2s + 4 − 2 t − 4 + 2 t − 3t = − 2 − 2s − 4t.
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3/15
Solution
x1 x2
s
x3
=
x4 x5
( ) )( ) ( )
19
3
5
1
− 8 + 4t
=
− 8
0
+
0
−4
t
0
s
19 5
− 4 + 2t
− 4t
− 2s
0
0
0
3
+
4
t
1 2
t
t
( ) ( ) () 1
−2
−2
0
19
− 8
5
−4
1
+s
0
−4 0 3
+t
fo r
=
1
−2
ut io n
() (
1
− 2 − 2s − 4t
D is tri b
9/9/2020
4
0
1
0
2
N ot
0
.
1
d. Because the system has infinitely many solutions, b = (4, 1, 5) T is a linear combination of the column vectors of A. →
iv. Processing math: 100%
a. The coefficient and the augmented matrices are
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4/15
9/9/2020
Solution
A=
(
−3 −2 −2
)| ( →
0
1 − 1 , (A b) = 0 0
0
−3 −2 −2 2 0
)
1 −1 2 . 0 0 4
0
ut io n
b. x 1, x 2 are basic variables and x 3 is a free variable. c. By the last equation of the system, we see that the system has no solutions. 2. 2. a. a)
D is tri b
↓ ( ) ( )↓ ( ) 1 −2 3 6
|
→
(A b) =
2
1
4 5
−3 1 −2 3
6
→
R1 ( 3 ) + R3
fo r
1 −2 3
R1 ( − 2 ) + R2
0 5 −2 −7
R2 ( 1 ) + R3
→
N ot
0 − 5 7 21 1 −2 3
6
0 5 −2 −7 . 0 0 5 14
Hence, we obtain Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
5/15
9/9/2020
Solution
x 1 − 2x 2 + 3x 3 = 6
(1)
5x 2 − 2x 3 = − 7
(2)
5x 3 = 14.
(3)
14
5x 2 = − 7 + 2x 3 = − 7 + 2
( ) 14 5
= −
7 5
,
7
D is tri b
and x 2 = − 25 . By (1),
ut io n
By (3), x 3 = 5 , by (2),
( ) ( )
x 1 = 6 + 2x 2 − 3x 3 = 6 + 2 −
−3
14 5
= −
74 . 25
N ot
fo r
b. b)
7 25
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6/15
Solution
(A b) =
( (
(
1
1 −2 3
2
3 −1 −4
−2 −3 −1 4
1 1 −2 0 1
3
3 − 10
0 − 1 − 5 10 1 1 −2
3
)
)↓
)↓
R1 ( − 2 ) + R2 R1 ( 2 ) + R3
R2 ( 1 ) + R3
→
D is tri b
|
→
ut io n
9/9/2020
0 1 3 − 10 . 0 0 −2 0
Hence, we obtain
fo r
x 1 + x 2 − 2x 3 = 3
(1) (2)
− 2x 3 = 0.
(3)
N ot
x 2 + 3x 3 = 10
By (3), x 3 = 0. By (2), x 2 = − 10 − 3x 3 = − 10. By (1), x 1 = 3 − x 2 + 2x 3 = 3 − ( − 10) = 13. Hence, (x 1, x 2, x 3) = (13, − 10, 0) is a solution. c. c)
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7/15
Solution
(A b) =
( (
(
1 2 3
4
4 7 6 17 2 5 12 7
1 2
3
4
0 −1 −6 1 0 1
6 −1
1 2
3 4
)↓ )↓
R1 ( − 4 ) + R2
→ R1 ( − 2 ) + R3
R2 ( 1 ) + R3
→
)
D is tri b
|
→
ut io n
9/9/2020
0 −1 −6 1 . 0 0 0 0
Hence, we have
(1)
− x 2 − 6x 3 = 1.
(2)
fo r
x 1 + 2x 2 + 3x 3 = 4
N ot
Let x 3 = t. By (2), x 2 = − 1 − 6x 3 = − 1 − 6t. By (1), x 1 = 4 − 2x 2 − 3x 3 = 4 − 2( − 1 − 6t) − 3t = 6 + 9t. Hence,
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8/15
9/9/2020
Solution
()( )()( )()() x2
6 + 9t
=
6
− 1 − 6t
=
−1
t
x3
9t
+
6
− 6t
0
−1
=
t
0
9
+ t −6 . 1
d. d)
(A b) =
3 −1 4 2 1 −1 5 −6 4 9
1 2 −1 3
Processing math: 100%
5
0 − 7 7 − 7 − 14
)↓ )↓ ( )
R1 ( − 3 ) + R2
14
1 2 −1 3
5
0 − 7 7 − 7 − 14 0 0
0
0
1 2 −1 3 5
→
R1 ( 1 ) + R3
R2 ( 1 ) + R3
fo r
0 7 −7 7
N ot
( ( (
(
2 −1 3 5
D is tri b
|
→
1
ut io n
x1
0
R2
→
1
−7
)
→
)
0 1 −1 1 2 . 0 0 0 0 0
Hence, we obtain
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9/15
9/9/2020
Solution
x 1 + 2x 2 − x 3 + 3x 4 = 5
(1)
x2 − x3 + x4 = 2
(2)
x 1 and x 2 are the basic variables and x 3 and x 4 are free variables. Let x 3 = s and x 4 = t, where s, t ∈ ℝ. By (2), we get x 2 = 2 + x 3 − x 4 = 2 + s − t and by (1),
ut io n
x 1 = 5 − 2x 2 + x 3 − 3x 4 = 5 − 2(2 + s − t) + s − 3t = 1 − s − t. We write the solution as a linear combination:
x2 x3
2+s−t s
=
t
2 0 0
−1
+s
1 1 0
−1
+t
−1 0
.
1
N ot
e. e)
=
1
fo r
x4
1−s−t
D is tri b
( ) ( ) () ( ) ( ) x1
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10/15
Solution
| (
R2
() 1 3
→ R3
() 1 2
2 −1 0 −1 4 −1
)↓
R1 ( − 2 ) + R2
→ R1 ( 1 ) + R3
( ) 1 −2 3
0 3 −6 0 2
2
↓ ( ) ( ) 1 −2 3
0 1 −2 0 1
R2 ( − 1 ) + R3
→
1
Hence, we have
0 1 −2 . 0 0 3
x 1 − 2x 2 = 3
0x 1 + x 2 = − 2 0x 1 + 0x 2 = 3
fo r
{
1 −2 3
D is tri b
→
(A b) =
1 −2 3
ut io n
9/9/2020
N ot
From the last equation, we see that the system has no solutions. f. f)
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11/15
Solution
| ( →
(A b) =
(
−1 2 −3 −4 2 −1 2 1
1 −1 2
−1 2 −3 −4 0 3 −4 −6 0 3 −4 −2
)↓
)↓
R2 ( − 1 ) + R3
→
→ R1 ( 1 ) + R3
(
−1 2 −3 −4
)
0 3 −4 −6 . 0 0 0 4
D is tri b
Hence, we obtain
− x 1 + 2x 2 − 3x 3 = − 4 0x 1 + 3x 2 − 4x 3 = − 6 0x 1 + 0x 2 + 0x 3 = 4
fo r
{
2
R1 ( 2 ) + R2
ut io n
9/9/2020
Section 4.3
N ot
From the last equation we see that the system has no solutions.
1. Solve each of the following systems using Gauss-Jordan elimination. a.
Processing math: 100%b.
{ {
x1 − x2 + x3 = 2 2x 1 − 3x 2 + x 3 = − 1 3x 1 − x 2 + 3x 3 = − 1 − 4x 1 + 2x 2 − 4x 3 = 2
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12/15
9/9/2020
Solution
f.
g.
ut io n
x 1 − 2x 2 + 2x 3 = 2 − 2x 1 + 3x 2 − 4x 3 = − 2 − 3x 1 + 4x 2 + 6x 3 = 0 x2 − x3 = 1
D is tri b
e.
− 3x 1 − x 2 − 2x 3 = 3
− x 1 + 3x 2 − x 3 = − 2 3x 1 + 3x 2 − 2x 3 = 4 x 1 + x 2 + 2x 3 = 1 − 2x 1 − 4x 2 − 5x 3 = − 1 3x 1 + 6x 2 + 5x 3 = 0 2x 2 + 3x 3 − 2x 4 = 3 2x 1 + 3x 3 + 2x 4 = 1 − 3x 1 + x 2 − 2x 3 = 3
fo r
d.
{ { { { {
4x 1 − 6x 2 + 6x 3 = 14
N ot
c.
x 1 + 3x 2 − x 3 = 2
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13/15
9/9/2020
Solution
{
− x 1 + 3x 2 − 4x 3 = − 3 x2 − x3 = − 2 x 1 − 2x 2 + 3x 3 = 1
ut io n
h.
x 1 − x 2 + 2x 3 = − 1
2. For each of the following homogeneous systems, solve it by using Gauss-Jordan elimination, write its solution as a linear combination, and find its solution space N A and dim(N A).
{
c.
d.
D is tri b
b.
2x − y + 3y = 0 2x − y + 3z = 0 4x − 2y + 6z = 0 − 6x + 3y − 9z = 0
{ {
x 1 + 4x 2 + 6x 3 = 0 2x 1 + 6x 2 + 3x 3 = 0 − 3x 1 + x 2 + 8x 3 = 0. − x 1 + 2x 2 + x 3 = 0 3x 1 − 3x 2 − 9x 3 = 0
fo r
{
N ot
a.
x + 2y − z = 0
− 2x 1 − x 2 + 12x 3 = 0.
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14/15
9/9/2020
Solution
x 1 + x 2 − 2x 3 − x 5 = 0
ut io n
x3 + x4 + x5 = 0
x 1 + x 2 − 2x 3 = 0 − 3x 1 + 2x 2 − x 3 = 0 − 2x 1 + 3x 2 − 3x 3 = 0 4x 1 − x 2 − x 3 = 0.
N ot
fo r
f.
{ {
− x 1 − x 2 + 2x 3 − 3x 4 + x 5 = 0
D is tri b
e.
2x 1 + 2x 2 − x 3 + x 5 = 0
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15/15
9/9/2020
Solution
Solution 1. 1.
|
(
)
(
R2 ( − 1 )
→
(
)
(
)
|
)
ut io n
1 − 1 1 2 R1 ( − 2 ) + R2 1 − 1 1 2 (A b) = → 2 −3 1 −1 0 −1 −1 −5 →
1 − 1 1 2 R2 ( 1 ) + R1 1 0 2 7 → = (B → c). 0 1 1 5 0 1 1 5
D is tri b
a.
The system of linear equations corresponding to (B | → c) is
x2 + x3 = 5
fo r
{
x 1 + 2x 3 = 7
x 2 = 5 − x 3 = 5 − t. Hence,
N ot
x 1 and x 2 are basic variables and x 3 is a free variable. Let x 3 = t. Then x 1 = 7 − 2x 3 = 7 − 2t and
( ) ( ) () ( ) () ( ) x1 x2 x3
− 7 − 2t
=
5−t t
7
=
5 0
− 5t
+
−t t
7
=
5 0
−5
+ t −1 . 1
b. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
1/21
9/9/2020
Solution
|
(
)
(
3 − 1 3 − 1 R2 ( 1 ) + R1 − 1 1 − 1 1 (A b) = → −4 2 −4 2 −4 2 −4 2 R1 ( − 1 )
→
( )
)
(
)
|
1
−2
→
(
)
(
)
1 − 1 1 − 1 R2 ( 1 ) + R1 1 0 1 0 → = (B → c). 0 1 0 1 0 1 0 1
D is tri b
R2
(
1 − 1 1 − 1 R1 ( 4 ) + R2 1 − 1 1 − 1 → −4 2 −4 2 0 −2 0 −2
)
ut io n
→
The system of linear equations corresponding to (B | → c) is
x2 = 1
fo r
{
x1 + x3 = 0
N ot
x 1 and x 2 are basic variables and x 3 is a free variable. Let x 3 = t. Then x 1 = − x 3 = − t and x 2 = 1. Hence,
( ) ( ) () ( ) () ( ) x1 x2 x3
−t
=
1 t
0
=
1 0
+
−t 0 t
0
=
1 0
−1
+t
0
.
1
c. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
2/21
Solution
| (
1
→
4 − 6 6 14
(A b) =
R1 ( − 4 ) + R2
→ R1 ( 3 ) + R3
( ) −
1 2
→
(
−3 −1 −2 3
(
3
−1 2
0 − 18 10 6 0
8
−5 9
) ( ) ( ) ( ) ( ) ( ) ( ) 1 3 −1 2
R3 ( − 1 ) + R2
0 9 −5 −3
→
0 8 −5 9 1 3 −1
R2 ( − 8 ) + R3
→
1
) )
2
R3
0 1 0 − 12
−
1 5
→
1 3 −1
9
2
0 1 0 − 12
fo r
0 0 1 − 21
1 3 0 − 19 R ( − 3 ) + R 1 0 0 17 2 1 0 1 0 − 12 0 1 0 − 12 . → 0 0 1 − 21 0 0 1 − 21
N ot
→
2
0 1 0 − 12 0 8 −5
0 0 − 5 105
R3 ( 1 ) + R1
1 3 −1
D is tri b
R2
3 −1 2
ut io n
9/9/2020
Hence, (x 1, x 2, x 3) = (7, − 12, − 21) is a solution of the system. d.
| ( →
(A b) =
1 −2 2
2
−2 3 −4 −2 −3 4
6
1 −2 2 2
0
)
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3/21
→
R1 ( 3 ) + R3
R2 ( − 1 )
→ R3
(
1
−2
( ) −6
→
0 − 2 12 6
0 1
)
)
2
0 −2
0 1 −6 −3
→
R3
0 −1 0 2
1 −2 2
R2 ( − 1 ) + R3
1
2 2 2
(
1 −2 2 0 1
2
0 −2
0 0 −6 −1
( ) 1 −2 2 2
0 1 0 −2
ut io n
( ()
R1 ( 2 ) + R2
Solution
1
)
D is tri b
9/9/2020
R3 ( − 2 ) + R1
1
→
fo r
0 0 1 6
( ) 5
1 −2 0 3
0 1 0 −2 1
0 0 1 6
N ot
( ) 7
R2 ( 2 ) + R1
→
1 0 0 −3
0 1 0 −2 . 1
0 0 1 6
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4/21
9/9/2020
Solution
(
7
1
Hence, (x 1, x 2, x 3) = − 3 , − 2, 6
)( ) ) () ( ) ( ) ( ) ( ) ( ) →
(A b) =
R1 ( 3 ) + R3
→
0 1 −1 1
−1 3 −1 −2 3 3 −2 4
1
1 7
→
0 0 7 − 14
0 1 −1 1 3 3 −2 4
−1 3 −1 −2 0 1 −1 1
0 0 1 −2
−1 3 0 −4 R ( −3) +R 2 1 0 1 0 −1 →
fo r
→
−1 3 −1 −2
0 12 − 5 − 2
0 1 −1
0 0 1 −2
N ot
R3 ( 1 ) + R1
→
2
− 1 3 − 1 − 2 R ( − 12 ) + R 2 3 0 1 −1 1 →
− 1 3 − 1 − 2 R3
R3 ( 1 ) + R2
R1 ,
ut io n
| ( (
is a solution of the system.
D is tri b
e.
)
−1 0 0 −1 R ( −1) 1 0 0 1 1 0 1 0 −1 0 1 0 −1 . → 0 0 1 −2 0 0 1 −2
Hence, (x 1, x 2, x 3) = (1, − 1, − 2) is a solution of the system. f. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
5/21
9/9/2020
Solution
R3 ( 1 ) + R2
→
R3
() 1 5
→
1
) ( ) ) ( )
2
1
−2 −4 −5 −1 3
6
5
1 1 2
0 1 −2 −2
→
0 3 −1 −3
( ) 1 1 2
R1 ( − 3 ) + R3
R2 ( − 3 ) + R3
3
0 0 1
5
1
0 3 −1 −3
1 1 2
1
0 1 −2 −2 0 0 5
3
( ) 1
1 1 0 −5
1
0 1 −2 −2
2
0 −2 −1 1
→
0
1
1 1
R1 ( 2 ) + R2
ut io n
(A b) =
1
D is tri b
| ( ( →
R3 ( 2 ) + R2
→
R3 ( − 2 ) + R1
4
0 1 0 −5 3
0 0 1 5
fo r
( )| 3
1 0 0 5
N ot
R2 ( − 1 ) + R1
→
4
0 1 0 −5
→
= (D d).
3
0 0 1 5
Hence, the solution is (x 1, x 2, x 3) =
(
3
4
3
)
, − 5, 5 . 5
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6/21
9/9/2020
Solution
| ( (
)(
0 2 3 −2 3 R 2 0 3 2 1 1, 2 2 0 3 2 1 → 0 2 3 −2 3 −3 1 −2 0 3 −3 1 −2 0 3
→
(A b) =
R3 ( 1 ) R1
)
−1 1 1
2 4 R ( −3) +R 1 3 0 2 3 −2 3 →
→
−3 1 −2 0 3
) ( ( ) ( ) () ( ) ( ( ) ( ) | 0
2
1
2
4
R2 ( 1 ) R3
3 −2 3
→
0 −2 −5 −6 −9
R3
−
1 2
→
−1 1 1
2
4
0 2 3 −2 3
)
D is tri b
(
−1 1
)
ut io n
g.
0 0 −2 −8 −6
−1 1 1 2 4 R ( −1) +R 3 1 0 2 3 −2 3 →
−1 1 0 −2
fo r
0 0 1 4 3 R3 ( − 3 ) + R2 1
0 2 0 − 14 − 6 4
R2 ( − 1 ) + R1
→
−1 1 0 −2 1
2
0 1 0 −7 −3
→
3
N ot
0 0 1
1
R2
−1 0 0 5
0 0 1 4
4
0 1 0 −7 −3 0 0 1 4
3
)
R1 ( − 1 )
→
3
1 0 0 −5 −1 0 1 0 −7 −3 0 0 1 4
= (B → c).
3
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7/21
(
9/9/2020
) |
Solution
The system of linear equations corresponding to (B | → c) is
x 2 − 7x 4 = − 3
ut io n
{
x 1 − 5x 4 = − 4 x 3 + 4x 4 = 3.
D is tri b
x 1, x 2 and x 3 are basic variables and x 4 is a free variable. Let x 4 = t. Then x 1 = − 4 + 5x 4 = − 4 + 5t, x 2 = − 3 + 7x 4 = − 3 + 7t, and x 3 = 3 − 4x 4 = 3 − 4t. Hence,
()( )()() x1
− 4 + 5t
x2
− 3 + 7t
h.
3 − 4t t
=
−3
7
3 0
+t
−4
.
1
N ot
x4
=
5
fo r
x3
−4
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8/21
9/9/2020
Solution
|( ) ( ) ( ) ( ) ( )| 1 −1 2 −1
(A b) =
− 1 3 − 4 − 3 R1 ( 1 ) + R2 → 0 1 −1 −2 R ( −1) +R 1 −2 3
1 −1 2 −1 0 2 −2 −4
R2
1 −1 2 −1
2
0 1 −1 −2 0 1 −1 −2
2
0 −1 1
2
D is tri b
0 −1 1
()
4
1
→
0 1 −1 −2
1
1
ut io n
→
1 −1 2 −1
R2 ( − 1 ) + R3
→
0 0
0
0
fo r
R2 ( 1 ) + R4
0 1 − 1 − 2 R2 ( 1 ) + R1 → 0 0 0 0
1 0 1 −3
= (B → c).
N ot
0 1 −1 −2 0 0 0
0
0 0 0
0
The system of linear equations corresponding to (B | → c) is
{
x1 + x3 = − 3 x 2 − x 3 = − 2.
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9/21
9/9/2020
Solution
x 1 and x 2 are basic variables and x 3 is a free variable. Let x 3 = t. Then x 1 = − 3 − x 4 = − 3 − t and x 2 = − 2 + x 3 = − 2 + t. Hence,
()( )()()()() −3 − t
x2
−3
−2 + t
=
−2
=
t
x3
−t
−3
t
+
0
−2
=
t
−1
+t
0
|
)
(
)
(
R2
1
)
−5
→
)
1 2 − 1 0 R2 ( − 2 ) + R1 1 0 1 0 → . 0 1 −1 0 0 1 −1 0
fo r
(
( )
D is tri b
(
1 2 − 1 0 R1 ( − 2 ) + R2 1 2 − 1 0 (A 0) = → 2 −1 3 0 0 −5 5 0 →
.
1
2. 2. a.
1
ut io n
x1
The system of linear equations corresponding to the last augmented matrix is
N ot
{
x+z=0 y−z=0
Let z = t. Then x = − t and y = t. We write the solution into a linear combination as follows:
() ( ) ( ) x y z
−t
=
t t
−1
=t
1
.
1
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10/21
9/9/2020
Solution
→ Let υ 1 = ( − 1, 1, 1) T. N A =
{
→ tυ1 : t ∈ ℝ
} {}
→ = span υ 1 . Because r(A) = 2, null(A) = 3 − 2 = 1 and
dim(N A) = 1.
| ( →
(A 0) =
) ( )
2 −1 3 0 R ( −2) +R 2 −1 3 0 1 2 4 −2 6 0 0 0 0 0 . → R1 ( 3 ) + R3 −6 3 −9 0 0 0 0 0
ut io n
b.
D is tri b
The system of linear equations corresponding to the last augmented matrix is 2x − y + 3z = 0. 1
3
Let y = s and t. Then x = 2 s − 2 t. We write the solution into a linear combination as follows:
y
→ Let υ 1 =
NA =
(
1
, 1, 0 2
{
)
T
3
s − 2t 2
) () ( ) 1
3
2
s t
=s 1 0
+t
−2 0
.
1
N ot
z
=
1
fo r
() ( x
(
)
→ T 3 and υ 2 = − 2 , 0, 1 . Then and
→ → s υ 1 + t υ 2 : s, t ∈ ℝ
} { }
→ → = span υ 1 , υ 2 . Because r(A) = 1, null(A) = 3 − 1 = 2 and
dim(N A) = 2. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
11/21
9/9/2020
Solution
| ( ) ( ) ( ) ( )( ) () ( ) ( ) ( ) ( ) 1 4 6 0 R ( −2) +R 1 4 6 0 1 2 2 6 3 0 0 −2 −9 0 →
→
(A 0) =
− 3 1 8 0 R1 ( 3 ) + R3
1
13
→
1 4
6 0 R 1 4 6 0 2, 3 0 −2 −9 0 → 0 1 2 0 0 1 2 0 0 −2 −9 0
R2 ( 2 ) + R3
→
1 4 6 0 R3 0 1 2 0 0 0 −5 0
R3 ( − 2 ) + R2
→
5
1 4 6 0
→
0 1 2 0 0 0 1 0
1 4 0 0 R ( −4) +R 1 0 0 0 2 1 0 1 0 0 0 1 0 0 . → 0 0 1 0 0 0 1 0
fo r
R3 ( − 6 ) + R1
1
D is tri b
R3
{ 0 } and dim(NA) = 0. →
N ot
Hence, x 1 = x 2 = x 3 = 0 is a solution of the system. N A = d.
0 13 26 0
ut io n
c.
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12/21
Solution
| (
) ( () ( ) ( ) ( ) ( ) →
(A 0) =
R1 ( − 1 ) , R2
→
R3
−
1
−1 2
1 0 R (3) +R 1 2 3 −3 −9 0 →
−1 2
− 2 − 1 12 0 R 1 ( − 2 ) + R 3 1 3
0
5
)
D is tri b
1 −2 −1 0 R (2) +R 1 0 −5 0 2 1 0 1 −2 0 0 1 −2 0 . → 0 0 0 0 0 0 0 0
3 −6 0
0 − 5 10 0
1 −2 −1 0 R ( −1) +R 2 3 0 1 −2 0 → 0 1 −2 0
1 0
ut io n
9/9/2020
The system corresponding to the last augmented matrix is
fo r
{
x 1 − 5x 3 = 0
x 2 − 2x 3 = 0.
N ot
Let x 3 = t. Then x 1 = 5t and x 2 = 2t. We write the solution into a linear combination as follows:
( ) ( ) () x1 x2 x3
5t
=
2t t
5
=t 2 . 1
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13/21
9/9/2020
Solution
→ Let υ 1 = (5, 2, 1) T. Then N A =
{
→ tυ1 : t ∈ ℝ
} {}
→ = span υ 1 . Because r(A) = 2, null(A) = 3 − 2 = 1
|( →
(A 0) =
2 −1 0
1 0
)
− 1 − 1 2 − 3 1 0 R1 , 2 → 1 1 −2 0 −1 0 0
0
1
1
−1 −1 2 −3 1 0
1 0
)
1
1 0 R1 ( 2 ) + R2 → 1 −2 0 −1 0 R (1) +R
0
0
2
2 −1 0 1
1
1 0
) )
1
3
() 1
fo r
−1 −1 2 −3 1 0
R2
3
0
0 3 −6 3 0
0
0 0 −3 0 0
0
0 1 1 1 0 R3 − 3
N ot
( ( (
2
D is tri b
e.
−1 −1 2 −3 1 0
→
( ) 1
0
0 1 − 2 1 0 R2 ( − 1 ) + R4 → 0 0 1 0 0
0
0 1 1 1 0
0
ut io n
and dim(N A) = 1.
−1 −1 2 −3 1 0 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
14/21
9/9/2020
( ) ( ) ( ) ( ) ( ) Solution
0
0 1 − 2 1 0 R3 ( − 3 ) + R4 → 0 0 1 0 0
0
0 0 3 0 0
0
0
0 1 − 2 1 0 R3 ( 2 ) + R2 → 0 0 1 0 0 R (3) +R
0
0 0 0 0 0
−1 −1 2 −3 1 0
3
1
D is tri b
0
0
0 1 0 1 0 R2 ( − 2 ) + R1 → 0 0 1 0 0
0
0 0 0 0 0
0
fo r
−1 −1 0 0 −1 0 0
0 1 0 1 0 R1 ( − 1 ) → 0 0 1 0 0
0
0 0 0 0 0
N ot
0
ut io n
−1 −1 2 −3 1 0
1 1 0 0 1 0 0 0 1 0 1 0 0 0 0 1 0 0
.
0 0 0 0 0 0
The system corresponding to the last augmented matrix is https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
15/21
9/9/2020
Solution
{
x1 + x2 + x5 = 0 x3 + x5 = 0 x 4 = 0.
ut io n
Let x 2 = s and x 5 = t. Then x 1 = − s − t, x 3 = − t and x 4 = 0. We can write the solution into a linear combination as follows: (x 1, x 2, x 3, x 4, x 5) T = s( − 1, 1, 0, 0, 0) T + t( − 1, 0, − 1, 0, 1) T.
NA =
D is tri b
→ → Let υ 1 = ( − 1, 1, 0, 0, 0) T and υ 2 = ( − 1, 0, − 1, 0, 1) T. Then
{
→ → s υ 1 + t υ 2 : s, t ∈ ℝ
} { }
→ → = span υ 1 , υ 2 .
N ot
f.
fo r
Because r(A) = 3, null(A) = 5 − 3 = 2 and dim(N A) = 2.
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16/21
9/9/2020
Solution
|( ) ( ) ( ) ( )
(A 0) =
− 3 2 − 1 0 R1 ( 2 ) + R3 → −2 3 −3 0 R ( −4) +R 1
4 −1 −1 0
1 1 −2 0
R2 ( − 1 ) + R3
→
R2 ( 1 ) + R4
0 5 −7 0
R2
0 0 0 0 0 0 0 0
( ) 3
1 0 −5 0
→
→
0 5 −7 0
4
0 5 −7 0 0 −5 7 0
1 1 −2 0 7
0 1 −5 0 0 0 0 0 0 0 0 0
7
fo r
R2 ( − 1 ) + R1
( ) 1
−5
1 1 −2 0
D is tri b
→
1 − 2 0 R1 ( 3 ) + R2
ut io n
1
0 1 −5 0 . 0 0 0 0
N ot
0 0 0 0
The system of linear equations corresponding to the last augmented matrix is
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9/9/2020
Solution
{
3
x1 − 5 x3 = 0 7
x2 − 5 x3 = 0
( ) ( ) () x2 x3
Section 4.4
7t 5t
=t 7 . 5
} {}
→ = span υ 1 . Because r(A) = 2, null(A) = 3 − 2 = 1
N ot
and dim(N A) = 1.
{
→ tυ1 : t ∈ ℝ
=
3
fo r
→ Let υ 1 = (3, 7, 5) T. Then N A =
3t
D is tri b
x1
ut io n
Let x 3 = 5t. Then x 1 = 3t and x 2 = 7t. We write the solution into a linear combination as follows:
1. Solve each of the following systems using the inverse matrix method. a.
{
3x 1 + 4x 2 = 7 x 1 − x 2 = 14
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18/21
9/9/2020
Solution
e.
{
ut io n
x 1 − 2x 2 + 2x 3 = 1 2x 1 + 3x 2 + 3x 3 = − 1 x 1 + 6x 3 = − 4 x 1 + 2x 2 + 3x 3 = 0
D is tri b
d.
x1 − x3 = 1
2x 1 + 5x 2 + 3x 3 = 1 x 1 + 8x 3 = − 1
x 1 − 2x 3 + x 4 = 0 x 2 + x 3 + 2x 4 = 1
fo r
c.
{ { {
2x 1 + 3x 2 + x 3 = 2
− x1 − x2 + x3 = − 1 2x 1 + 9x 4 = 1
N ot
b.
x1 + x2 + x3 = 2
2. Show that if a ≠ 3, then the following system has a unique solution for (b 1, b 2) ∈ ℝ 2.
{
x − y = b1 ax − 3y = b 2
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Solution
3. For each (b 1, b 2, b 3) ∈ ℝ 3, show that the following system has a unique solution.
− 3x 1 + 4x 2 + 6x 3 = b 2 − x 1 − 2x 2 + 3x 3 = b 3
4. Solve each of the following systems by using Cramer’s rule.
c.
d.
{ { {
D is tri b
2x + 5y = − 1 x 1 + 2x 3 = 2 − 2x 1 + 3x 2 + x 3 = 16 − x 1 − 2x 2 + 4x 3 = 5 x1 + x2 − x3 = 1 2x 1 + x 2 − x 3 = 0 x 2 − 2x 3 = − 2 x1 + x2 + x3 = 0 2x 1 − x 2 − x 3 = 1
fo r
b.
{
N ot
a.
x − 3y = 6
ut io n
{
x 1 + 2x 3 = b 1
x 2 + 2x 3 = 2
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9/9/2020
Solution
{
x2 + x3 = 0 x1 − x2 = 0 x3 − x4 = 1
N ot
fo r
D is tri b
ut io n
e.
x1 − x4 = 0
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9/9/2020
Solution
Solution 1. 1.
|
3
|
4
(
1 −1 = − 7, A − 1 = − 7 −1 −1
1
−4 3
)
. By Theorem 4.4.1, the solution of the
system is
() () ( = A −1
x2
( R3
14
2
3
1
0
1
0 R ( −2) +R 1 2 0 →
1
0
−1
0
0
1 R1 ( − 1 ) + R3
0
1
−1
−2
1
0
−1
−2
−1
0
1
1
1
−1
(
14
0
0
→
)
)( ) ( )
1
1
3
3
1
1
( )
7 −1
7
1
1
1
−4
1
1
−
1 −1
1 0 0
fo r
| (
(A I) =
= −
0
1
9
−5
.
) (
1
1
1
0
0
1
−1
−2
1
0
1
0
−3
−3
1
1
1
0
0
−2
1
0
1
=
0 R (1) +R 1 2 3 0 0 →
N ot
b.
7
D is tri b
x1
ut io n
a. Because |A | =
1
−3
1
−3
)
0
)
R3 ( 1 ) + R2
→ R3 ( − 1 ) + R1
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) )|
Solution
1
0
0
1
0
0
0
1
1
0
0
0
1
0
0
0
1
1
1
3
3
2
−1
−3
3 1
1
−3
1
−3
1
2 3
2
1
= (I 3 A − 1).
−3
3 1
1
→
−3
1
−1
R2 ( − 1 ) + R1
1
1
−3
−3
()
( ) 1
2
−3
N ot
x1
fo r
By Theorem 4.4.1, the solution of the system is
1
x2
=
x3
c. Processing math: 100%
| (
(A I) =
2
−1 1
ut io n
1
0
D is tri b
( (
9/9/2020
3
1
−3
3
1
−3 1
−3
1
−2
2
1
0
2
3
3
0
1
() ( ) 2 2 1
)
2
=
−1 . 1
0 R ( −2) +R 1 2 0 →
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6
0
)
1
1(
)
2
1
0
0
7
−1
−2
1
0 R ( −3) +R 3 2 0 →
0
2
4
−1
0
1
1
−2
2
1
0
0
0
1
− 13
1
1
−3
0
2
4
−1
0
1
1
−2
2
1
0
0
1
− 13
1
1
0
0
30
−3
−2
0
− 24
0
1
− 13
0
0
1
3
2
1
1
(
)
R2 ( − 2 ) + R3
→
)()
0
R2 ( 2 ) + R1
D is tri b
1
1
− 10
−3
→
7
R3
−6 −3
1
− 15
7
− 30
3
2 5
−5
2
1
15
30
1
7
1
0
0
5
0
1
0
0
0
1
− 10
3 1
− 10
3
)
−2
=
Processing math: 100%
0
1
N ot
(
Solution
0
fo r
( ( (
1
ut io n
| (
9/9/2020
− 15
2
30
)
1
30
R 3 ( 13 ) + R 2
→ R 3 ( 24 ) + R 1
)|
= (I 3 A − 1).
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Solution
By Theorem 4.4.1, the solution of the system is
x2 x3
=
( ) () 2
5
5
−5
2
1
15
30
1
7
3
− 10 1
− 10
− 15
()
9 5
1
−1 −4
30
=
17
− 30 29
− 30
.
N ot
fo r
D is tri b
d.
2
ut io n
() x1
3
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9/9/2020
Solution
1
0
2
5
3
0
1
0 R ( −2) +R 1 2 0 →
1
0
8
0
0
1 R1 ( − 1 ) + R3
)
1
2
3
1
0
0
1
−3
−2
1
0 R (2) +R 2 3 0 →
0
−2
5
−1
0
1
1
2
3
1
0
0
1
−3
−2
0
0
−1
−5
(
3
1
1
−3
−2
2
1
0
1
5
1
2
0
− 14
6
0
1
0
13
−5
0
0
1
5
−2
1
0
0
− 40
16
0
1
0
13
−5
0
0
1
5
−2
0
0
1
0
−2
−1
D is tri b
→ R3 ( − 3 ) + R1
) (
2
1
0 R ( −1) 1 3 0 0 → 0
3
−3 −1
)
)
R2 ( − 2 ) + R1
→
) |
fo r
R3 ( 3 ) + R2
(
3
9
−3 −1
= (I 3 A − 1).
N ot
( (
)
2
ut io n
| (
(A I 3) =
1
By Theorem 4.4.1, the solution of the system is
()( x1 x2
Processing math: 100%
x3
=
− 40
16
9
13
−5
3
5
−2
−1
)( ) ( ) 0 1
−1
7
=
−2 . −1
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Solution
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−1
1
1
0
0
0
0
1
1
2
0
1
0
−1
−1
1
0
0
0
1
0 R1 ( 1 ) + R3 → 0 R ( −2) +R
2
0
0
9
0
0
0
1
1
)
0
−1
1
1
0
0
0
0
1
1
2
0
1
0
0
−1
0
1
1
0
1
0 R2 ( 1 ) + R3 → 0
0
0
2
7
−2
0
0
1
4
D is tri b
1
0
)
1
0
−1
1
1
0
0
0
1
1
2
0
1
0
0
0
1
3
1
1
1
0
0
2
7
−2
0
0
1
0
−1
1
1
0
0
0
0
1
1
2
0
1
0
0
0
0 R4 ( − 3 ) + R3 → 0 R ( −2) +R ,
0
0
N ot
( ( ( (
)
0
ut io n
|(
(A I 4) =
1
1
0 R3 ( − 2 ) + R4 → 0 1
fo r
e.
)
1
3
1
1
1
0
1
−4
−2
−2
0
−1
0
5
2
2
−1
0
1
1
0
8
5
4
0
0
1
0
13
7
7
− 2 R3 ( − 1 ) + R2 → −3 R (1) +R
0
0
0
1
−4
−2
−2
1
1
4
2
R4 ( − 1 ) + R1
)
3
1
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(
Solution
1
0
0
0
18
9
9
−4
0
1
0
0
−5
−2
−3
1
0
0
1
0
13
7
7
−3
0
0
0
1
−4
−2
−2
1
)|
= (I 4 A − 1).
By Theorem 4.4.1, the solution of the system is
x2 x3
=
x4
Because |A | =
| | 1 −1 a −3
9
9
−4
−5
−2
−3
1
13
7
7
−4
−2
−2
4
1
=
2
−3 1
−1 1
−3
.
2
= − 3 + a, | A | ≠ 0 if a ≠ 3. By Corollary 4.4.1, the system has a unique
A=
(
(
1
N ot
solution for each (b 1, b 2) ∈ ℝ 2. 3. 3.
Processing math: 100%
)( ) ( ) 0
fo r
2. 2.
18
D is tri b
()( x1
ut io n
9/9/2020
0
−3
4
−1
−2
1
0
0
2
0
−2
) ( ) ) ( ) 2 R (1) +R 1 1 3 6 0 →
0
2
4
12
R1 ( 3 ) + R2
−2
5
3
0
2 R (1) +R 1 2 3 6 0 →
0
2
2
5
0
6 . 11
0
R2
() 1 2
→
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Solution
Hence, r(A) = 3. By Corollary 4.4.1, the system has a unique solution for each (b 1, b 2, b 3) ∈ ℝ 3. 4. 4. a. Because −3
2
5
1
6
2
−1
| |
= 11,
| A1 | =
x=
13
, − 11 11
)
5
|
27 11
and y =
is the solution of the system.
= 27,
|A 2| |A|
= −
13 11
.
N ot
b.
(
27
−1
fo r
Hence,
|A|
=
−3
= − 13.
This implies |A 1|
|
6
ut io n
|A 2 | =
| |
1
D is tri b
|A | =
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9/9/2020
Solution
|A 3| =
3
1
−1
−2
4
2
0
2
16
3
1
5
−2
4
1
2
2
−2
16
1
−1
5
4
1
0
2
−2
3
16
−1
−2
5
This implies
|A1 |
0
−2
|
3 = (12 + 8) − ( − 6 − 2) = 28.
−1
−2
2
0
16
3 = (24 − 64) − (30 − 4) = − 66.
5
−2
1
2
−2
16 = (64 − 2 − 20) − ( − 32 + 5 − 16) = 85.
−1
5
1 −2 −1
66 33 x1 = = − = − , |A| 28 14
(
33
85
61
Hence, − 14 , 28 , 28
)
ut io n
−2
1
0
D is tri b
|A 2| =
| | |
2
3 = (15 + 8) − ( − 6 − 32) = 61. −2
fo r
|A 1| =
| | | |
0
N ot
|A| =
1
|A2 |
85 x2 = = |A| 28
and x 3 =
|A3 | |A|
=
61 . 28
is the solution of the system.
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9/16
9/9/2020
Solution
|A 3| =
2
1
−1
0
1
−2
1
1
−1
0
1
−1
−2
1
−2
1
1
−1
2
0
−1
0
−2
−2
1
1
1
2
1
0
0
1
−2
This implies
|A|
=
−1
1
2
1 = ( − 2 − 2) − ( − 1 − 4) = 1,
0
1
| |
|
1
1
0
1 = ( − 2 + 2) − (2 − 1) = − 1,
−2
1
1
1
2
0 = 4 − (2 − 4) = 6,
0
−2
1
1
2
1 = ( − 2 + 2) − ( − 4) = 4.
0
1
= − 1,
x2 =
|A 2|
N ot
x1 =
|A 1|
|
1
1
ut io n
|A 2| =
−1
D is tri b
|A 1| =
| | | |
1
fo r
|A| =
1
|A|
=
6
|A 3|
4 = 6 and x 3 = = = 4. 1 |A| 1
Hence, ( − 1, 6, 4) is the solution of the system.
d. The coefficient matrix is A =
(
1
1
2
−1
1
0
1
)
1 . Then 2
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Solution
1
−1
2
0
−1 , 2
1
1
0
2
−1
1
0
| |=
− 3,
A3 =
By computation, we obtain A 1
8
7
x 1 = 3 , x 2 = − 3 , and x 3 = 3 . Hence,
A2 =
1
0
2
1
1
2
)
1 . 2
| A1 | =
(
1
8
− 1, 7
, − 3, 3 3
)
− 1 , and 2
| A2 | = 8, | A3 | =
− 7. By (4.4.3),
)
is a solution of the system.
N ot
fo r
e.
1
ut io n
A1 =
1
) (
1
D is tri b
( (
0
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11/16
Solution
|A 2 | =
|A 3 | =
|A 4 | = Processing math: 100%
−1
0
1
1
0
1
−1
0
0
0
0
1
−1
0
0
0
−1
1
1
1
0
0
−1
0
0
1
0
1
1
1
0
0
−1
0
1
1
0
1
0
0
0
0
1
1
−1
|
1
0
0
−1
0
1
1
0
1
−1
0
0
0
0
1
−1
1
0
0
0
0
1
1
1
1
−1
0
0
0
0
1
1
|
||
1
= −1 0
1
0
0
0
1
||
0
− 1 −1 0
1 −1 0
1
|
0 = − 2. 1
ut io n
| | | |
0
= 0 because columns 1 and 3 are the same.
= 0 because columns 2 and 3 are the same.
|
fo r
|A 1 | =
|
0
= |A| = − 2
N ot
|A | =
1
D is tri b
9/9/2020
|
= 0 because columns 3 and 4 are the same.
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Solution
| A1 |
| A2 |
| A3 |
−2
Hence, x 1 = | A | = 0, x 2 = | A | = 0, x 3 = | A | = − 2 = 1, x 4 = 0, and (0, 0, 1, 0) is a solution of the system.
Section 4.5
c.
d.
{
2x 1 + x 2 + 4x 3 = 5
D is tri b
− 3x 1 + x 2 − 2x 3 = 3 x 1 + x 2 − 2x 3 = 3 2x 1 + 3x 2 − x 3 = − 4 − 2x 1 − 3x 2x 3 = 4 x 1 + 2x 2 + 3x 3 = 4 4x 1 + 7x 2 + 6x 3 = 17 2x 1 + 5x 2 + 12x 3 = 7 x 1 − 2x 2 − x 3 + 2x 4 = 0
fo r
b.
{ { {
x 1 − 2x 2 + 3x 3 = 6
N ot
a.
ut io n
1. For each of the following systems, determine whether it has a unique solution, infinitely many solutions, or no solutions.
− x 2 − 2x 3 + x 4 = 0
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Solution
f.
{ {
− 2x 1 + 2x 2 − 6x 3 = 5 2x 1 − 2x 2 + 6x 3 = 4 x 1 − x 2 − 3x 3 − 4x 4 = 1 − x1 + x2 − x3 = − 1
b.
{ {
2x 1 − 3x 2 − x 3 − 5x 4 = − 1 − 3x 1 + 6x 2 − 2x 3 + x 4 = 3 x1 + x4 = − 2 x 1 − 3x 2 + 2x 3 = − 1 x 1 + 4x 2 − x 3 = 2
fo r
a.
D is tri b
2. For each of the following systems, determine if it has a unique solution.
ut io n
e.
x 1 − x 2 + 3x 3 = 2
a.
{
x 1 − 2x 2 − x 3 = 0
N ot
3. For each of the following homogeneous systems, determine whether it has a unique solution or infinitely many solutions.
− x 1 + 3x 2 + 2x 3 = 0 − x1 + x2 = 0 x 2 + x 3 = 0.
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Solution
d.
{
{
− x 1 − 2x 2 − 3x 3 = 0. x1 − x2 + x3 = 0
ut io n
c.
{
2x 1 − x 2 + 4x 3 = 0
2x 2 − 4x 3 + x 3 = 0.
x 1 − x 2 + 2x 3 = 0 − 2x 1 − 2x 2 + x 3 = 0
D is tri b
b.
x 1 + x 2 − 2x 3 = 0
− x 1 − 3x 2 + 3x 3 = 0 3x 1 + x 2 + x 3 = 0.
6x 1 − 4x 2 = b 1 3x 1 − 2x 2 = b 2
N ot
{
fo r
4. Consider the following system
1. Find b 1 and b 2 such that the system is consistent. 2. Find b 1 and b 2 such that the system is inconsistent. →
3. Determine whether the system with b = (4, 1) T is inconsistent. 5. Determine if the following system is consistent for all b 1, b 2, b 3 ∈ ℝ. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
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Solution
− 3x 1 + 4x 3 − x 4 = b 3
6. Find conditions on b 1, b 2, b 3 such that the following system is consistent.
D is tri b
{
x 1 − x 2 + 2x 3 = b 1 2x 1 + 4x 2 − 4x 3 = b 2
ut io n
{
x 1 − x 2 + 2x 3 + x 4 = b 1 2x 1 + 4x 2 − 4x 3 = b 2
− 3x 1 − 2x 3 = b 3
7. Determine whether the following system has infinitely many solutions for all b 1, b 2 ∈ ℝ.
fo r
− 3x 1 + 3x 2 + x 3 = b 2
N ot
{
x 1 − 2x 2 + 2x 3 = b 1
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Solution
Solution 1. 1. a.
| (
1
→
2
(A b) =
−3
R2 ( 1 ) + R3
→
|) ( | ) ( |) −2
3
6
1
4
5
1
−2 3
1 −2 0
5
0
0
3
R1 ( − 2 ) + R2
→
R1 ( 3 ) + R3
1 −2 0
5
0 −5
3
6
−2 −7 7
21
6
−2 −7 . 5 14
→
Hence, r(A) = r(A | b) = 3. By Theorem 4.5.1 (1), the system has a unique solution. b.
| ( →
(A b) =
|)
1
1
−2 3
2
3
−1 −4
−2 −3 −1 4
R1 ( − 2 ) + R2
→
R1 ( 2 ) + R3
( | ) ( | ) 1
1
−2
6
0
1
3
− 10
0 −1 −5 −2
→
R2 ( 1 ) + R3
→
1 1 −2 0 1
6
3
− 10 . 0 0 − 2 − 12
Hence, r(A) = r(A | b) = 3. By Theorem 4.5.1 (1), the system has a unique solution. c. …
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Solution
| ( |) ( |) ( |) →
(A b) =
1 2
3
4
4 7
6 17
R1 ( − 4 ) + R2
→
R1 ( − 2 ) + R3
2 5 12 7
1
R2 ( 1 ) + R3
2
1
2
3
4
0 −1 −6 1 0
1
6 −1
3 4
0 −1 6 1 . 0 0 0 0
→
→
Hence, r(A) = r(A | b) = 2 < 3, . By Theorem 4.5.1 (2), the system has infinitely many solutions. →
→
d. Because n = 4 and m = 2, r(A) = r(A | 0) = r(A | b) ≤ m < n. By Theorem 4.5.1 (2), the system has infinitely many solutions. e.
(
→
(A | b) =
|
1
−1
−2
2
2
|) ( |)
3
2
R ( 2 ) + R2
−6 5
−2
6
4
1 −2 3 2
→
0
0
R1 ( − 2 ) + R3
0
0
0 9 . 0 0
→
Hence, r(A) = 1 and r(A b) = 2. By Theorem 4.5.1 (3), the system has no solutions. f.
|
→
(A b) =
( …
2/9
(
1
| ) |)
−1 −3 −4 1
−1
1
−1
0 −1
1 −1 −3 −4 1 0
0
−4 −4 0
.
R1 ( 1 ) + R2
→ R1 ( − 2 ) + R3
9/9/2020
Solution →
Hence, r(A) = r(A | b) = 2 < 4 = n. By Theorem 4.5.1 (2), the system has infinitely many solutions. 2. 2.
io n
a. Because m = 3 < 4 = n, by Corollary 4.5.2, the system doesn’t have a unique solution b. Because m = 3 < 3 = n, by Corollary 4.5.2, the system doesn’t have a unique solution 3. 3.
( ) ( ) ( ) 3
2
−1
1
0
0
1
1
R1 ( 1 ) + R2
→
R1 ( 1 ) + R3
0
1
rib
−1
1 −2 −1 1
0 −1 −1
is t
A=
−2 −1
0
1
1
rD
1
ut
a. Because
1 −2 −1
→
1
1
0
0
0
0
0
0
ot
R2 ( − 1 ) + R4
0
fo
R2 ( 1 ) + R2
,
N
r(A) = 2. Note that m = 4 and n = 3. Hence, n = 3 < 4 = m and r(A) = 2 < n and by Theorem 4.5.2 (2), the system has infinitely many solutions. b. Because
…
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Solution
|A | =
|
1
1
2
−1
|
−2 1 4
1
2
− 1 = (3 − 4 + 8) − ( − 2 − 8 − 6) = 23 ≠ 0, −1 −2 −3 −1 −2
( ) ( ) ( ) 1 −1
2
−2 −2 1
R1 ( 1 ) + R3
0 −4
5
0 −4
5
−1 −3 3 3
1
1
→
R1 ( − 3 ) + R4
rib
R1 ( 2 ) + R2
0
4
−5
rD
A=
−1 2
is t
1
ut
io n
it follows from Theorem 4.5.3 that the system has a unique solution. c. Because m = 2 < 3 = n, it follows from Theorem 4.5.2 (ii) that the system has infinitely many solutions. d. Because
1 −1 2
→
0
0
0
0
0
0
ot
R2 ( 1 ) + R4
0 −4 5
fo
R2 ( − 1 ) + R3
,
N
r(A) = 2. Hence, n = 3 < 4 = m and r(A) = 2 < 3 = n. It follows from Theorem 4.5.2 (2) that the system has infinitely many solutions. 4. 4. 1. If b 2 −
b1 2
→
= 0, then r(A) = r(A | b) = 1 < 2. It follows from Theorem 4.5.1 (2) that the system has
infinitely many solutions and is consistent. …
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Solution
2. If b 2 −
b1 2
→
→
≠ 0, then r(A) = 1 and r(A | b) = 2. Hence, r(A) < r(A | b). By Theorem 4.5.1 (3), the
system has no solutions and is inconsistent. →
3. By b = (4, 1) T, we have b 1 = 4 and b 1 ≠ 2b 2, it follows from result (2) that the system is inconsistent.
2
4
−4
0
0
4
−1
−3
1
→
1 −1 0
6
0
0
2
→
R1 ( 3 ) + R3
1
−8 0 . 6 2
1 −1 0
6
2
1
−8 0
0 − 3 10 2
rD
R2 ( 2 ) + R3
R1 ( − 2 ) + R2
)
tio
) ( ( ) 1
bu
(
2
tri
=
−1
is
A
1
n
5. 5.
Because r(A) = 3 = m, by Theorem 4.5.4 (1), the system is consistent for any b 1, b 2, b 3.
N
ot
fo
6. 6.
…
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Solution
→
(A | b)
( |) ( | ) =
1
−1
2
4
−3
0
1 −1
2
0
2
b1
R1 ( − 2 ) + R2
− 4 b2
− 2 b3
b1
− 8 b 2 − 2b 1
6
0 −3
4
→
R1 ( 3 ) + R3
1
R2 ( 2 ) + R3
→
b 3 + 3b 1
( | ) 1 −1 0 0
6
b1
2
b 2 − 2b 1
−8
.
1
0
0 b + (b + 2b ) 3 1 2 2
1
If b 3 + 2 (b 2 + 2b 1) = 0, then the system is consistent 7. 7.
A=
(
1 −1 2
4
2 −4
)
R1 ( − 2 ) + R2
→
(
1 −1 0
6
2 −8
)
.
Because r(A) = 2 = m and m = 2 < 3 = n, by Theorem 4.5.5, the system has infinitely many solutions for any b 1, b 2 ∈ ℝ.
Section 4.6 …
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Solution
1. Let S =
{
→ → → → a 1, a 2, a 3, a 4
}
→ → → → and A = (a 1, a 2, a 3, a 4), where
→ a1 =
() ( ) () ( ) () 1
2 , 3
→ a2 =
−2 1
−1
,
→ a3 =
3
4 , 7
→ a1 =
−1
−4 , −5
2
→
b=
2 . 4
1. Determine whether AX = b is consistent, where X = (x 1, x 2, x 3, x 4) T. →
→
→
→
2. Determine whether the vector b is a linear combination of S. →
3. Determine whether the vector b ∈ span S. → → → → 2. Let a 1 = (1, − 2, 2) T, a 2 = ( − 1, 0, − 10) T, a 3 = (1, − 1, 6) T, and b = (b 1, b 2, b 3) T.
{ { {
} } }
→ → → → 1. Find conditions on b 1, b 2, b 3 such that b ∈ span a 1, a 2, a 3 . → → → 2. Find conditions on b 1, b 2, b 3 such that b ∉ span a 1, a 2, a 3 . →
→ → → → → 3. Find a vector b = (b 1, b 2, b 3) T such that b ∉ span a 1, a 2, a 3 .
…
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Solution
3. Let S =
{( ) ( )} 2 1
−4
,
−2
.
1. Determine whether span S = ℝ 2. →
2. Determine whether b = (4, 2) T is in span S. 3. Determine whether b = (6, 1) T is in span S. →
4. Determine whether span S = ℝ 2, where i. S =
ii. S =
iii. S =
{( ) ( ) ( ) ( )} {( ) ( ) ( )} {( ) ( ) ( ) ( )} −1 0
1
−1 0 1
,
1
,
0
2
,
−2
2 1
1
,
,
−2
3
,
5
−5
,
.
5
3
−2
,
.
4 1
.
5. Determine whether span S = ℝ 3, where
S=
6. Determine whether span S = ℝ 4, where
…
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{( ) ( ) ( )} 1
2
3
2 , 1
5 , 0
4 5
.
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Solution
S=
…
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{( ) ( ) ( )} 1
1
1
0
0
2
4
,
0
,
3
.
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Solution
Solution →
⎛
→
1
(A| b ) = ⎜ 2 ⎝
⎛
1
−2
⎜0 ⎝
0
Let
−5 ∣ 4
−1 ∣ ∣
2
−2
−2 ∣ −2
b
…
1/4
⎠
R1 (−3)+R3
R2 (−1)+R3
⎛
1
⎠
⎝
0
−2 5 0
3
−1 ∣
−2
−2
0
0
∣ ∣
2
is consistent.
→→→
and A = (a1 a2 a3 ).
⎞
−2 ⎟ .
∣
By Theorem 4.6.1, we have the following conclusions.
∈ span S.
= (b1 , b2 , b3 )
R1 (−2)+R2
2⎟ − −−−− →
−−−− → ⎜0 −2 ⎟ −
5
∣
→ b
⎞
is a linear combination of S.
→ b
7
N
2. 2.
−1
∣
⎞
fo
3. The vector
b
−4
ot
2. The vector
→
=
4
−2
→
→
∣
1
−2
b ) = 2.
1. The system A X
3
−1 ∣ 2
3
5
→
Hence, r(A) = r(A|
3
−2
io n
b ).
bu t
We find r(A) and r(A|
rD is tri
1. 1.
0
⎠
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Solution 1
−1
(A| b ) = ⎜ −2
⎛
1
1∣
−1
⎜0
2
∣
−10
6
b1
−8
4 ∣ b3 − 2b1
1
−1
⎜0
−2
1∣ ∣ 1∣
0 1
= ⎜0 ⎝
0
1∣
−1 −2
1
R1 (−2)+R3
⎞
⎞
⎟.
b2 + 2b1
∣
0 ∣ b3 − 4b2 − 10b1
0
2. If b3
− 4b2 − 10b1 ≠ 0,
then r(A) < (A|
b ).
→
By Theorem 4.6.1,
fo
Then, b3
b ).
⎠
By Theorem 4.6.1,
rD
→
then r(A) = (A|
= 1.
⎠
⎠
b1
∣
− 4b2 − 10b1 = 0,
and b3
∣ b3
R2 (−4)+R3
b1
1. If b3
= b2 = 0
R1 (2)+R2
−−−− → b2 ⎟ −
⎟ b2 + 2b1 ∣ ⎠ 0 ∣ b3 − 2b1 − 4(b2 + 2b1 )
0
⎛
∣
∣
⎞
−−−− → b2 + 2b1 ⎟ −
0
⎝
3. Let b1
⎞
1
⎛
− 4b2 − 10b1 ≠ 0.
→ b → b
→ → → ∈ span{a1 , a2 , a3 }. → → → ∈ span{a1 , a2 , a3 }.
Hence,
ot
→ → → = (0, 0, 1) ∉ span{a1 , a2 , a3 }.
N
b
−1
−2
⎝
→
0
∣
n
⎝
∣ b1
1
is tri bu tio
⎛
→
3. 3. 1. Because
2
−4
A = (
…
2/4
4 2
)
1
−2
) −− → ( 1
r(A) = 1 < 2 = m.
R1 (
−2
By Theorem 4.6.2, span S
R1 (−1)+R2
1
−2
0
0
) − −−−− → ( 1
−2
2
≠ R .
),
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Solution
(A| b )
= (
R1 ( ) −4 ∣ b1 1 2 ) −− → ( ∣ −2 ∣ b2 1
2 1
2. 1
−2
0
0
(
1. If b2
−
→ b
2
→
≠ 0,
then r(A) = (A| then b2
= (b1 , b2 ) = (4, 2),
2. If b2
−
→ b
b1
b1 2
b )
then r(A) < (A| then b2
= (b1 , b2 ) = (6, 1)
2
b )
−
b1 2
b1 ∣ 2 ∣ ∣ ∣ b2 −
b1
).
2
→
and by Theorem 4.6.1,
b1
− →
≠ 0,
b1 R1 (−1)+R2 −2 ∣ 2 ) − −−−− → ∣ −2 ∣ b 2
1
→
= 2−
4 2
and
= 0
b
b
6 2
Hence, if
= (4, 2) ∈ span S. →
and by Theorem 4.6.1, = 1−
∈ span S.
→
b
∉ span S.
and
= −2 ≠ 0
Hence, if
→ b
= (6, 1) ∉ span S.
4. 4. i. Let A = (
−1
1
1
3
0
0
−2
5
r(A) = 2 = m < n = 4, 2
−5
−1
−2
5
Theorem 4.6.2, span S
⎛
5. 5.
1
3
1
1
−2
A = ⎜2 ⎝
1
5 0
3
⎞
).
⎠
2
−5
0
0
0
2
= R .
R1 (−2)+R2
⎛
1
R1 (−1)+R3
⎝
0
2 1 −2
3
1
2
− − − − →⎜0 −2 ⎟ −
1
5
⎞
R2 (2)+R3
⎠
Hence, r(A) = 3. By Theorem 4.6.2, span S = R3 . Because m = 4 and n = 3, by Theorem 4.6.2, span S 6. 6. …
3/4
) , r(A) = 1 < 2 = m.
By
Then m = 2 and n = 4. Because r(A) = 2 = m < n = 3, by
−−−− → ⎜0 4⎟ − 5
1
2
2
2
R1 (1)+R2
) − − − − − →(
2
= R .
≠ R .
0
Theorem 4.6.2, span S
Then m = 2 and n = 4. Because
by Theorem 4.6.2, span S
1
ii. Because A = (
iii. Let A = (
).
⎛
⎝
4
≠ R .
0
0
3
⎞
−2 ⎟ . 1
⎠
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…
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Solution
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A.5 Linear transformations
A.5 Linear transformations Section 5.1 1. For each of the following matrices,
⎝
⎛
−1
2
3
0
0
1
1
−2
0
3
1
−2
3
2
−1
⎜ −2 A3 = ⎜ ⎜ 3 ⎝
4
1
2
−1
⎞
⎛
⎟
A2 = ⎜
⎠
⎝
2
1
1
−3 ⎟
1
0
−2
−1
1
⎜ 2 A4 = ⎜ ⎜ −1
0
⎛
⎞ ⎟ ⎟ ⎟
−2
⎝
⎠
determine the domain and codomain of TA
i
1
⎞
⎠
⎞
−2 ⎟ ⎟ 1 ⎟
−1
0
1
0
1
⎠
, i = 1, 2, 3, 4.
t
A1 = ⎜ −1
−2
is
2
D
⎛
Moreover, compute
TA (0, 1, 1, 2), TA (1, −2, 1), TA (−1, 0, 1), TA (−1, 1, 1).
2. Let A = ( 3. Let A = (
3
2
−1
3
−1
2
1
3
0
6
).
Find TA ( ).
1
4
)
0 →
and TA (
0
or
2
1
)
1
.
ot f
3
→
→
Find TA ( e1 ), TA ( e2 ), TA ( e3 ), where
→ → → e1 , e2 , e3
are the standard vectors in
. 4. For each of the following linear transformations, express it in (5.1.1) . a. T (x1 , x2 ) = (x1 − x2 , 2x1 + x2 , 5x1 − 3x2 ). b. T (x1 , x2 , x3 , x4 ) = (6x1 − x2 − x3 + x4 , x2 + x3 − 2x4 , 3x3 + x4 , 5x4 ). c. T (x1 , x2 , x3 ) = (4x1 − x2 + 2x3 , 3x1 − x3 , 2x1 + x2 + 5x3 ). d. T (x1 , x2 , x3 ) = (8x1 − x1 + x3 , x2 − x3 , 2x1 + x2 , x2 + 3x3 ). 3
5. Let …
1/2
N
R
→
→
X = (2, −2, −2, 1), Y
= (1, −2, 2, 3)
and
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A.5 Linear transformations ⎛ A = ⎜ ⎝
→
1
0
1
0
1
−2
1
1⎟.
−1
1
2
0
⎞
⎠
→
Compute TA ( X − 3 Y ) . 6. Let S, T : R3 → R3 be linear operators such that T (1, 2, 3) = (1, −4) and S(1, 2, 3) = (4, 9). Compute (5T + 2S)(1, 2, 3) . 7. Let S, T
be linear transformations such that T (1, 2, 3) = (1, 1, 0, −1) and S(1, 2, 3) = (2, −1, 2, 2). Compute (3T − 2S)(1, 2, 3) . 8. Compute (T + 2S)(x1 , x2 , x3 , x4 ), where 3
: R
4
→ R
T (x1 , x2 , x3 , x4 ) = (x1 − x2 + x3 , x1 − x2 + 2x3 − x4 ),
is
t
S(x1 , x2 , x3 , x4 ) = (x2 − x3 + 2x4 , x1 + x3 − x4 ).
N
ot f
or
D
9. For each pair of linear transformations TA and TB , find TB TA . a. TA (x1 , x2 ) = (x1 + x2 , x1 − x2 ), TB (x1 , x2 ) = (x1 − x2 , 2x1 − 3x2 , 3x1 + x2 ). b. TA (x1 , x2 , x3 ) = (−x1 − x2 + x3 , x1 + x2 ), TB (x1 , x2 ) = (x1 − x2 , 2x1 + 3x2 ). c. TA (x1 , x2 , x3 ) = (−x1 + x2 − x3 , x1 − 2x2 , x3 ), TB (x1 , x2 , x3 ) = (x1 − x2 + x3 , 2x1
…
2/2
+ 3x2 + x3 ).
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Solution
Solution 1. 1.
Because the size of A 1 is 3 × 4, the domain and codomain of T A are ℝ 4 and ℝ 3, respectively. 1
() ( 0
TA
1
1
1
=
2
−1 4 −1
−1
2
3
0
1
2 −1
2
0
)( ) ( ) 0
0
1
5 . 1
=
1 2
Because the size of A 2 is 3 × 3, the domain and codomain of T A are ℝ 3. 2
( ) ( )( ) ( ) 1
TA
2
−2
−2 2
=
1
1
1
1
1 −3
−2
1
0 −2
1
−5
=
−4 . −1
Because the size of A 3 is 4 × 3, the domain and codomain of T A are ℝ 3 and ℝ 4, respectively. 3
()( −1
TA
3
0 1
=
1
−2
0
−2
3
1
3
−2
3
1
2
−1
)( ) ( −1 0
=
1
−1 3 0 −2
)
.
Because the size of A 4 is 4 × 3, the domain and codomain of T A are ℝ 3 and ℝ 4, respectively. 4
…
1/6
9/9/2020
Solution
()( −1
TA
4
1
=
1
−1
1
1
2
0
−2
−1 −1 0
1
0
1
)( ) ( −1 1
=
1
3 −4 1 1
2. 2.
T
T
3. 3.
4. 4. …
2/6
T(→ e 1) =
(
−1 2 1
T(→ e 2) =
(
−1 2 1
T(→ e 3) =
(
−1 2 1
3
3
3
0 6
0 6
0 6
( ) ( )( ) ( ) ( ) ( )( ) ( ) 1
=
0
0
=
1
)( ) ( ) 1 0
=
0
−1 3
)( ) ( ) 0 1
=
0
2 0
,
)( ) ( ) 0 0 1
=
1 6
.
,
1
2
1
−1 3
0
1
2
0
−1 3
1
=
=
1
,
−1 2 3
.
)
.
9/9/2020
Solution
a. T
() x1 x2
=
1
5 −3
x2 x3
=
1
x1
1
1
−2
x2
0
0
3
1
x3
0
0
0
5
x4
x2
=
x1
d. T x 2 x3
=
x1
2
3
0
−1
2
1
5
8
3/6
.
0
4 −1
x3
…
x2
( ) ( )( ) ( ) ( )( ) x1
5. 5.
)( ) x1
6 −1 −1
x4
c.
2
( ) ( )( ) x1
b. T
(
1 −1
x2 . x3
0
0
−1 0
1
x1
0
1 −1
x2 .
2
1
0
x3
0
1
3
By Proposition 5.1.1, we have
.
9/9/2020
Solution →
→
→
→
T(X + 3Y) = T(X) + 3T(Y)
( )( () ()() 1
=
0
1
1 0
1
−2 1 1
1
−1
=
2 0
1
3
6
+ 3 10 1
−6
6. 6.
1
() () () 1
1
1
(5T + 2S) 2 3
= 5T 2 3
+ 2S 2 3
=5
+3
0
1 0
−2 1 1
−1
1
2 0
−4
−2 2 3
)
10
=
36 . −3
( ) () ( ) 1
)(
1
+2
4 9
=
13
−2
.
7. 7. (3T − 2S)(1, 2, 3) = 3T(1, 2, 3) − 2S(1, 2, 3) = 3(1, 1, 0, − 1) − 2(2, − 1, 2, 2) = ( − 1, 5, − 4, − 7). 8. 8. Let → a = (T + 2S)(x 1, x 2, x 3, x 4). By Definition 5.1.3, we have a = (x 1 − x 2 + x 3, x 1 − x 2 + 2x 3 − x 4) + 2(x 2 − x 3 + 2x 4, x 1 + x 3 − x 4)
→
= (x 1 + x 2 − x 3 + 4x 4, 3x 1 − x 2 + 4x 3 − 3x 4). 9. 9. a. Because
…
4/6
9/9/2020
Solution
() x1
(ST)
() x1
(ST) x 2
=
x3
( )(
x2
() x1
(ST) x 2
=
x3
)
()
1
(
−2 −2 1
(
1 −1 1
(
1
2 3
1
1
1
2 3
2
1
)
)(
−2 3 1
0
−4 −1
)( ) (
3
−1 −1 1 0
1
1
1 −1
x1 x2
)
=
(
0
2
−1 5 4
2
)( ) x1 x2
x1 x2 x3
() x1
x2 . x3
−1 1
−1
1
−2 0
0
0
)
()
c.
=
2 −3
1 −1
b.
=
(
=
1 −1
1
)( ) x1 x2 x3
x1
x2 . x3
Section 5.2 1. For each of the following linear transformations, determine whether it is onto. …
5/6
.
9/9/2020
Solution
a. T(x 1, x 2, x 3) = (2x 1 − 2x 2 + x 3, 3x 1 − 2x 2 + x 3, x 1 + 2x 2 − 4x 3, 2x 1 − 3x 2 + 2x 3). b. T(x 1, x 2, x 3) = (x 1 + 2x 2 − x 3, − x 1 − 2x 2, − x 1 + x 2 + 2x 3). c. T(x 1, x 2, x 3, x 4) = (x 1 − 2x 2 + x 4, 3x 1 + 2x 4, x 1 − 2x 3 + 3x 4). d. T(x 1, x 2, x 3, x 4) = (x 1 − x 2 + x 3, x 1 − x 4, 2x 1 − x 2 + x 3 − x 4).
n
2. Let T : ℝ 2 → ℝ 2 be defined by
bu tio
T(x, y) = (x − y, 2x − 2y).
3. Show that T is not onto and find a vector b ∈ ℝ 2 such that b ∉ T(ℝ 2). 4. For each of the following linear transformations, determine whether it is one to one. a. T(x 1, x 2, x 3) = (x 1 − x 2 + 2x 3, − x 1 − x 2, x 1 + x 2 − x 3, − 2x 1 − x 2 + x 3). →
b. T(x 1, x 2, x 3) = ( − x 1 + x 2 + x 3, x 1 − x 2 − 2x 3, + x 1 + x 2).
rD
d. T(x 1, x 2, x 3) = (x 1 − x 2 + 3x 3, x 1 − 2x 3, 2x 1 − x 2 + x 3).
is
c. T(x 1, x 2, x 3, x 4) = (x 1 − x 2 + x 3, 2x 1 + x 2 + x 3, x 1 + x 2 + 2x 4).
tri
→
b. T(x 1, x 2) = (3x 1 − 9x 2, 4x 1 − 12x 2).
ot
c. T(x 1, x 2, x 3) = (x 1 + 3x 2 + 2x 3, x 2 − x 3, x 3).
fo
5. For each of the following linear operators, determine whether it is invertible. If so, find its inverse. a. T(x 1, x 2) = (x 1 + x 2, 6x 1 + 7x 2).
N
d. T(x 1, x 2, x 3) = (x 1 + 3x 2, 3x 1 + 7x 2 − 4x 3, x 1 + 5x 2 + 5x 3).
…
6/6
9/9/2020
Solution
Solution 1. 1.
|
−1
b. Because A = − 1 − 2 −1 1
0 = 3 ≠ 0, by Theorem 5.2.5, T is onto. 2
) ( ) (
2
0
−2 3
1 −2 0
1
0 2
−2 2
0 6
0
−1
N
1/7
0 6
−1
0 2
R2 ( − 3 ) + R3
r(A) = 3 = m. By Theorem 5.2.3, T is onto. d. Because
…
→
R1 ( − 1 ) + R3
fo
1 0
1
→
)
is
3 0
1 −2 0
R1 ( − 3 ) + R2
ot
(
(
1
0
rD
A =
1 −2 0
tri
c. Because
tio
2
bu
|
1
n
a. The standard matrix A of T is a 4 × 3 matrix, By Theorem 5.2.3, T is not onto because r(A) ≤ 3 = min{4, 3} < 4 = m.
1 −2 0
−2 2 1
0 2
−2 2
0 0
6
−7
)
,
R2 , 3
→
Solution
(
1 0
) ( ( ) 0 −1
2 −1 1 −1
R2 ( − 1 ) + R3
→
0 1
| | 1 −1 2 −2
0 1
−1 −1
R1 ( − 2 ) + R3
0 1
−1 −1
−1 −1 , 0 0
= 0, by Theorem 5.2.5, T is not onto. Consider the system
rD
2x − 2y = b 2.
R1 ( − 2 ) + R2
→
( |
1 − 1 b1 0
0
N
ot
2 − 2 b2
is
tri ( | ) 1 − 1 b1
x − y = b1
fo
{
If b 2 − 2b 1 ≠ 0, the system has no solutions. We choose →
b = (b 1, b 2) = (1, 1).
a. Because …
2/7
)
0
(A.5.1)
3. 3.
0
bu
Because |A | =
1 −1 1
→
1 −1 1 0 0
2. 2.
R1 ( − 1 ) + R2
n
A =
1 −1 1 0
tio
9/9/2020
b 2 − 2b 1
)
.
9/9/2020
Solution
( ) ( ) ( ) ( ) 1
−1 2
→ R 1( − 1) + R 3 R 1(2) + R 4
−1 −1 0
A =
1
1
−1
−2 −1 1
1 −1 2
R1 ( 1 ) + R3
→
R2
( ) −
3
2
+ R4
1 −1 2
R1 ( 1 ) + R2
0 −2 2 0 0
−1
0 0
2
0 −2 2 0 2
−3
0 −3 5
1 −1 2
R3 ( 2 ) + R3
→
0 −2 2 0 0
−1
0 0
0
,
r(A) = 3 = n. By Theorem 5.2.4, T is one to one.
| | | | −1
b. Because |A | =
1 1
1
1
− 1 − 2 = − 2 ≠ 0, by Theorem 5.2.5, T is one to one. 1 0
c. The standard matrix of T is a 3 × 4 matrix. By Theorem 5.2.4, T is not one to one. 1 −1
3
d. Because |A | = 1 0 − 2 = 0, by Theorem 5.2.5, T is not one to one. 2 −1 1 4. 4. a. Because |A | =
…
3/7
| | 1 1 6 7
= 1 ≠ 0, T is invertible. Because
9/9/2020
Solution
|A|
T −1
b. Because |A | =
|
3
4 − 12
() x1 x2
1
7
−1
−6
1
7
−1
x1
−6
1
x2
,
.
1
(A I) =
2
) ( ( ) | ( ) ( )( ) 1 0 0
R3 ( 1 ) + R2
0 1 −1 0 1 0 0 0
1
R2 ( − 3 ) + R1
→
0 0 1
1 3 0 1 0 −2
→
0 1 0 0 1
1
R3 ( − 2 ) + R1
0 0 1 0 0
1
1 0 0 1 −3 −5 0 1 0 0
1
1
0 0 1 0
0
1
x
T y z
4/7
=
) ( ) ( )( ) =
≠ 0, T is invertible. Because
1 3
…
−6
−1
2
0 1 −1 0 0
7
= 0, by Theorem 5.2.5, T is not invertible.
| | | ( 1 3
c. Because |A | =
|
−9
(
1
A −1 =
=
= (I 3 A − 1),
1 −3 −5
x
0
1
1
0
0
1
y . z
)
9/9/2020
Solution
| | | ( ( 3 7 −4
3 7 −4 0 1 0 1 5
R2 ( 1 ) + R3
→
1
1 3 0 0 1 2
…
5/7
0
1
0
1 3 2
) ( ( ) ) →
R1 ( − 1 ) + R3
0 0
R2
0
0
1
−2 0 1
0
) (
1 0 0 − 0 1 0
0 0 1
55
15
2
2
19 2
−4
6
5
− 2 −2 1
1
1
0 0
0 −2 −4 −3 1 0 0
2
5
−1 0 1
1
2
→
R3 ( − 2 ) + R2
→
3
1 3 0
1
0 1 0
19 2
0
0 0 1 −4
)|
0
5
− 2 −2
fo
1
−
1
−4 1 1
ot
(
1
N
→
0 0 1
R1 ( − 3 ) + R2
0 −2 −4 −3 1 0
0 0 1 −4
R2 ( − 3 ) + R1
5 3
0
(
1 0 0
)
tio
(A I) =
0
bu
1 3
= (I 3 A − 1).
1
n
5
tri
1 5
= − 2 ≠ 0, T is invertible.
is
d. Because |A | =
0
rD
1 3
1
)
9/9/2020
Solution
() x
=
55
15
2
2
19 2
−4
6
5
− 2 −2
1
1
x
y . z
bu tio
n
Hence, T y z
( )( ) −
Section 5.3 1. In Theorem 5.3.1
π π
, if θ = 6 , 3 , find T
( ) 4
−4
.
tri
2. Find the linear operator T = T 2T 1 on ℝ 2, where T 1 is a dilation with factor k = 2 on ℝ 2 and T 2 is a 1. Find the linear operator on ℝ 2 that contracts with factor
1
2
and then reflects about the line y = x.
rD
3.
is
projection on the line y = x.
2. Find the linear operator on ℝ 2 that reflects about the y-axis and then reflects about the x-axis,
fo
and the linear operator on ℝ 2 that reflects about the x-axis and then reflects about the y-axis.
π
4
and the image of x =
ot
4. Find the rotation operator with the rotation angle of
N
5. Find the linear operator on ℝ 2 that rotates by 6. Find the linear operator on ℝ 3 that rotates by yz-plane. 7. Prove Theorem 5.3.1 8. Prove Theorem 5.3.2 9. Prove Theorem 5.3.2
…
6/7
π
3 π 4
( ) 1
−1
under the operator.
radians about the origin. radians about the z-axis and then reflects about the
by using Theorem 5.1.1 . by using a method similar to the proof of Theorem 5.3.1 by using Theorem 5.1.1 .
.
Solution
N
ot
fo r
D
is t
rib
ut
io n
9/9/2020
…
7/7
9/9/2020
Solution
Solution
=
( ) 4
−4
=
sin
sin
6
( )( ) (
3
√3
4
4
4
√3
1
−4
4
4
cos 2 cos
π
π
sin
3
sin
3
π 6
)
( ) (√ ) 4
−4
π
π
3
cos
sin 2
3
=
…
1/8
( ) 1
√3
4
4
√3
3
π 3
π
3
)( ) 4
−4
4
4
( ) (√ ) 4
−4
=
3 − √3 3−1
.
tio
6
2
6
bu
cos
π
6
π
N
T
(
π
cos
tri
=
6
π
is
−4
sin
rD
( )
π
fo
T
4
cos 2
n
By Theorem 5.3.1,
ot
1. 1.
=
1 + √3 3+3
.
9/9/2020
Solution
By Definition 5.1.2, T 1(x, y) = (2x, 2y) for (x, y) ∈ ℝ 2 and by Corollary 5.3.1 (3),
(
x+y 2
,
x+y 2
)
. It follows that
y
0
0
1
2
( ) ( ) ( )( )
= T 2T 1
x y
,
T2
y
()
N
0
=
1
2
4. 4. π
1. By (5.3.6) and θ = 4 ,
x
0 1
=
y
() ( ) x
=
0 1 1 0
ot
y
2
bu
=
() x
() 1
x
.
tri
() x
T
2/8
= (x + y, x + y).
1 0
y
is
T1
…
)
By Definition 5.1.2 (2), we have
( )( )
fo
3. 3.
(
2x + 2y 2x + 2y , 2 2
tio
T(x, y) = T 2(T 1(x, y)) = T 2(2x, 2y) =
n
T 2(x, y) =
rD
2. 2.
1
2
0
() x y
.
1
2
0
0
1
2
x y
9/9/2020
Solution
T
( ) 1
−1
( ) ( )( ) cos
=
sin
=
π
4
π
− sin
cos
4
π
4
π
4
( ) 1
−1
√2
=
2
−
√2 2
√2
√2
2
2
1
−1
( ) √2 0
.
2. Because the reflection operator about the y-axis is T 1(x, y) = ( − x, y) and the reflection operator about the x-axis is T 2(x, y) = (x, − y), we have T 1T 2(x, y) = T 1(T 2(x, y)) = T 1(x, − y) = ( − x, − y) and T 2T 1(x, y) = T 2( − x, y) = ( − x, − y). 5. 5.
…
3/8
9/9/2020
Solution
T
() ( x
=
y
cos θ − sin θ sin θ
cos θ
( ) 1
=
2
−
√3
1
2
2
√3 2
)( ) x y
( )( ) π
cos
=
− sin
3
π
sin
cos
3
π
x
3
y
π
3
() x y
.
6. 6.
() ( x
T1 y z
=
cos θ − sin θ 0 sin θ
cos θ
0
0
0
1
y z
( )( ) ( ) ( ) π
π
√2
π
cos 4 − sin 4 0
=
)( ) x
π
sin 4
cos 4
0
0
0
1
x y
2
=
z
T2 y z …
4/8
=
2
√2
√2
2
2
0
( ) ( )( ) x
−
√2
0
−1 0 0
x
0
1 0
0
0 1
y and z
0
x
0
y . z
1
Solution
( )( ) ( )( ) √2
0
0 1
2
2
2
0
0
x
0
y . z
2
√2
2
2
0
0
√2
√2
0
7. 7.
−
0
1
y z
1
n
√2
x
tio
1 0
√2
0
2
0
−
=
−
tri
=
2
rD
T 2T 1 y z
−1 0 0
√2
is
() ( ) x
√2
bu
9/9/2020
T(→ e1) = T(1, 0) T = (b 1, b 2) T. Then b 1 = r cos θ and b 2 = r sin θ (see Figure 6 (I)). Because
fo
r = cosθ, b 1 = cos 2 θ, b 2 = cosθ sin θ. Hence,
T
ot
T(→ e 1) = (cos 2 θ, cos θ sin θ) . Let T(→ e 2) = T(0, 1) T = (u 1, u 2) T. Then u 1 = r cos θ and u 2 = r sin θ (see Figure 6 (II)). Because
N
r = sinθ, u 1 = sinθ cos θ, and u 2 = sin 2 θ,
T
T(→ e 2) = (sinθ cos θ, sin 2 θ) . By Theorem 5.1.1, the standard matrix for T is
…
5/8
9/9/2020
Solution
(T(→ e 1), T(→ e 2)) =
(
cos 2θ
sinθcosθ
cosθsinθ
sin 2θ
)
and (5.3.1) holds. Figure A.1: (I), (II)
8. 8.
() ( w1 w2
=
=
and
…
6/8
(
r cos(θ + β) r sin(θ + β)
) ( )( ) =
r cos β cos θ − r sin β sin θ r cos β sin θ + r sin β cos θ
cosθ − sinθ
r cos β
sinθ
r sin β
cosθ
)
9/9/2020
Solution
() ( ( x y
=
=
r cos(β − θ) r sin(β − θ) cosθ sinθ sinθ
) ( )( )
r cos β cos θ + r sin β sin θ
=
− cosθ
r cos β sin θ − r sin β cos θ
r cos β
)
.
r sin β
Solving the above equation, we have
( ) ( r cos β r sin β
=
cos θ
sin θ − cos θ
( ) ( ( w1 w2
sin θ
=
sin θ
sin θ
=
(
=
(
cos θ
=
y
cos θ − sin θ
cos θ − sin θ
=
) () ( x
−1
cos θ
)(
cos θ
sin θ − cos θ
)( ) r cos β r sin β
cos θ
sin θ
sin θ − cos θ
)( ) )( )
cos 2 θ − sin2 θ
2 sinθ cos θ
2 sinθ cos θ
sin 2 θ − cos 2 θ
cos 2θ
sin 2θ
sin 2θ − cos 2θ
sin θ
x y
x y
)( ) x y
and (5.3.5) holds. By the first figure in 8, since OB = OA = 1, we have 9. 9. (w 1, w 2) = T(1, 0) = (cos 2θ, sin 2θ).
…
7/8
)( ) x y
.
9/9/2020
…
8/8
Solution
9/9/2020
A.6 Lines and planes in Math Equation
A.6 Lines and planes in R3
2. Let → a ,
→ → → ( a × b ) ⊥ b
and
→ a = (1, 2, 3); b
= (1, 0, 1),
2.
→ a = (1, 0, − 3), b
3.
→ → a = (0, 2, 1); b = (−5, 0, 1),
4.
→ → a = (1, 1, − 1); b = (−1, 1, 0),
→ a = (1, − 2, 3), →
→ b , c
= (−1, 0, − 2),
→ b
= (−1, − 4, 0),
and
.
2.
→ → a = (1, 0, 0), b = (−1, 1, − 2).
Find the scalar triple product of
and
→ b
.
fo
→ → a = (1, 0, 1), b = (−1, 0, 1);
→ c = (2, 3, 1).
N
ot
1.
→ a
4. Find the volume of the parallelpiped determined by
1/1
ut
→
3. Find the area of the parallelogram determined by
…
.
→
is tri b
1.
and verify
→ → → ( a × b ) ⊥ a
rD
1. Find
→ → a × b
io n
Section 5.4
→
1.
→ a = (1, − 1, 0), b
2.
→ a = (−1, − 1, 0),
= (−1, 0, 2),
→ → a , b,
→ c
.
→ c = (0, − 1, − 1).
→ b
and
= (−1, 1, − 2),
→ c = (−1, 1, 1).
9/9/2020
Solution
Solution 1. 1. →
1. → a×b=
(| | | | | |) 2 3 0 1
1 3
, −
1 2
,
1 1
= (2, 2, − 2).
1 0
Because →
(→ a × b) ⋅ → a = (2)(1) + (2)(2) + ( − 2)(3) = 0, and →
(→ a × b) ⋅ → a = (2)(1) + (2)(0) + ( − 2)(1) = 0, →
→
→
(→ a × b) ⊥ → a and (→ a × b) ⊥ b. →
2. (→ a × b) =
(| | | 0 −3 0 −2
, −
1
−3
−1 −2
| | |) ,
1
0
−1 0
= (0, 5, 0).
Because →
(→ a × b) ⋅ → a = (0)(1) + (5)(0) + (0)( − 3) = 0 and →
(→ a × b) ⋅ → a = (0)( − 1) + (5)(0) + (0)( − 2) = 0, →
→
→
(→ a × b) ⊥ → a and (→ a × b) ⊥ b. →
3. → a×b= …
1/4
(| | | 2 1 0 1
, −
0
1
−5 1
| | |) ,
0
2
−5 0
= (2, − 5, 10).
9/9/2020
Solution
Because →
(→ a × b) ⋅ → a = (2)(0) + ( − 5)(2) + (10)(1) = 0 and →
(→ a × b) ⋅ → a = (2)( − 5) + ( − 5)(0) + (10)(1) = 0, →
→
→
(→ a × b) ⊥ → a and (→ a × b) ⊥ b. →
4. → a×b=
(| | | 1 −1 1
0
, −
1
−1
−1
0
| | |) 1
,
1
−1 1
= (1, 1, 2).
Because →
(→ a × b) ⋅ → a = (1)(1) + (1)(1) + (2)( − 1) = 0 and →
(→ a × b) ⋅ → a = (1)( − 1) + (1)(1) + (2)(0) = 0, →
→
→
(→ a × b) ⊥ → a and (→ a × b) ⊥ b. 2. 2.
By (6.1.3), we have
→
→
a×b=
(| | | −2 3 −4 0
, −
1
3
−1 0
|| ,
1
−2
−1 −4
|)
Hence, →
→
a ⋅ ( b) × → c) = 24 − 9 − 6 = 9.
…
2/4
= (12, − 3, − 6).
9/9/2020
Solution
3. 3. 1. Because →
→
a×b=
(| | | 0 1 0 1
, −
1
1
−1 1
| | |) ,
1
0
−1 0
= (0, − 2, 0),
by Theorem 6.1.1, the area equals
∥
→
→
a×b∥ =
√02 + 22 + 12 = √5.
4. 4.
| |
1
−1
0
2 1. Because − 1 0 0 −1 −1 −1 −1
2. Because − 1 −1
0
1
−2
1
1
| |
= 3, V = | 3 | = 3.
= − 6, V = | 1 − 6 | = 6.
Section 5.5 1.
a. Find the parametric equation of the line passing through the point P 0( − 1, 2, 0) that is parallel →
to the vector υ = (1, − 1, 1). b. Where does the line intersect the xy-plane? 2. Find the parametric equation of the line that passes through P 1(2, 1, − 1) and P 2( − 2, 3, − 5). 3. Find the distance from the point Q( − 1, 2, 1) to the line …
3/4
9/9/2020
Solution
x = 1 + t, y = 2 − 4t, z = − 3 + 2t. 4. Find the distance from the point Q( − 1, 2) to the line y = − 2x + 4.
…
4/4
9/9/2020
Solution
Solution 1. 1. a. By (6.2.1), x = − 1 + t, y = 2 − t, z = t. b. When the line intersects the xy-plane z = 0 and t = 0. When t = 0, x = − 1 and y = 2. Hence, the intersection point is ( − 1, 2, 0).
2. 2.
→ Let P 0 = P 1(2, 1, − 1) and v = P 1P 2 = ( − 4, 2, − 4). By (6.2.1), →
x = 2 − 4t,
3. 3.
y = 1 + 2t,
z = − 1 − 4t.
→ Take P(1, 2, − 3) on the line. Then PQ = ( − 2, 0, 4). Let → v = (1, − 4, 2). Then → v is parallel to the
line. By computation,
∥v ∥
=
→
→ v × PQ =
→
√21 and
(| | | −4 2 0
∥ ∥
, −
4
1
2
−2 4
|| ,
1
−4
−2
0
|)
= ( − 16, − 8, − 8),
→
v × PQ
→
By Theorem 6.2.2, D = 4. 4.
∥v ∥ →
1/3
8√14 7
.
Rewrite y = − 2x + 4 as 2x + y = 4.
D=
…
=
|Ax1 + By1 − C | √
A2 + B2
=
|2( − 1) + 2 − 4|
√
( − 1) 2 + 2 2
=
4√5 5
.
9/9/2020
Solution
Section 5.6 1. Find a normal vector for each of the following planes. a. x + 2y − z = 1, b. 4x − 2y − 3z = 5, c. − x − 6y + z − 1 = 0, d. 2x + 3z − 4 = 0. 2. Determine whether the following planes are parallel or orthogonal. a. x − y + z = 2, 2x − 2y + 2z = 5, b. 2x − y + z = 1, x + y − z = 6, c. 2x − y + 3z = 2, x + 2y + z = 4, d. 4x − 2y + 6z = 3, − 2x + y − 3z = 5 3. For each pair of the following planes, find a parametric equation of the line of intersection of the two planes. a. x − y + z = 2, 2x − 3y + z = − 1; b. 3 x − y + 3z = − 1, − 4x + 2y − 4z = 2, c. x − 3y − 2z = 4, 3x + 2y − 4z = 6. 4. Find an equation for the line through P(2, − 3, 0) that is parallel to the planes 2x + 2y + z = 2 and x − 3y = 5. 5. Find an equation for the line through P(2, − 3, 0) that is perpendicular to the two lines x = − 1 + y, y = 2 − 2t, z = 1 − t and x = 1 − t, y = 1 − t, z = 2 + t. 6. Find an equation of the plane passing through the point P and perpendicular to the vector → n. a. P(1, − 1, 0), → n = (1, 3, 5) → b. P(1, − 1, 3), n = ( − 1, 0, − 1) c. P(0, − 1, 2), → n = ( − 1, 1, − 1) d. P(1, 0, 0), → n = (1, 1, − 1) …
2/3
9/9/2020
Solution
7. Find an equation for the plane through P(2, 7, − 1) that is parallel to the plane 4x − y + 3z = 3. 8. Find an equation of the plane passing through the following given three points: a. P 1(1, 2, − 1), P 2(2, 3, 1), P 3(3, − 1, 2) b. P 1(1, 0, 1) P 2(0, 1, − 1) P 3(1, 1, − 2)
is t
x = 2 − 3t, y = 1 + t, z = 2t.
rib
ut
io n
9. Find an equation for the plane containing the line x = 3 + 6t, y = 4, z = t and that is parallel to the line of intersection of the planes 2x + y + z = 1 and x − 2y + 3z = 2. 10. Find an equation for the plane through P(1, 4, 4) that contains the line of intersection of the planes x − y + 3z = 5 and 2x + 2y + 7z = 0. 11. Find an equation of the plane passing through the point Q(1, 2, − 1) and perpendicular to the planes x + y + z = 2 and − x + 2y + 3z = 5. 12. Find an equation for the plane through P(1, 1, 3) that is perpendicular to the line
N
ot
fo
rD
13. Find an equation for the plane that contains the line x = 3 + t, y = 5, z = 5 + 2t, and is perpendicular to the plane x + y + z = 4. 14. Find the distance from the point to the plane. a. P(1, − 2, − 1), x − y + 2z = − 1; b. P(0, 1, 2), 2x + y + 3z = 4; c. P(1, 0, 1), 2x − 5y + z = 3; d. P( − 1, − 1, 1), − x + 2y + 3z = 1. 15. Find the distance between the given parallel planes. a. x − y + 2z = − 3, 3x − 3y + 6z = 1 b. x + y − z = 1, 2x + 2y − 2z = − 5
…
3/3
9/9/2020
Solution
Solution 1. 1.
By (6.3.1), we have n = (1, 2, − 1), a. → → b. n = (4, − 2, − 3), n = ( − 1, − 6, 1), and c. → d. → n = (2, 0, 3).
2. 2. → → → → → → a. Let n 1 = (1, − 1, 1) and n 2 = (2, − 2, 2). Then n 1 = 2n 1 and n 1 ∥ n 2. → → → → → → b. Let n 1 = (2, − 1, 1) and n 2 = (1, 1, − 1). Then n 1 ⋅ n 2 = 0 and n 1 ⊥ n 2. → → → → c. Let n 1 = (2, − 1, 3) and n 2 = (1, 2, 1). Then n 1 is not parallel to n 2 because they are not → → → → proportional. Because n 1 ⋅ n 2 = 3 ≠ 0, n 2 is not orthogonal to n 1. → → → → → → d. n 1 = (4, − 2, 6) and n 2 = ( − 2, 1, − 3). Then n 1 = − 2n 2 and n 1 ∥ n 2. 3. 3.
…
1/7
For a) and b), see the solutions to the exercises a) and b) in Section 4.3.
9/9/2020
Solution
| ( →
c) (A b) =
1
R 2( 11 )→
1 −3 −2 4 3
(
2
−4 6
)
1 −3 −2 0
4
2
1
R 1( − 3) + R 2
6
− 11
11
(
)
1 −3 −2 0 11
R 2(2) + R 1→
(
4
2
−6
)
16
26
1 0 − 11 0 1
2 11
11 6
− 11
)
.
The system corresponding to the last augmented matrix is
{
x− y+
16
z= 11
26 11
2
z= − 11
6 11
,
where x and y are basic variables and z is a free variable. Let z = t. Then x = y= −
6 11
−
2
z= − 11
6 11
−
2 11
t Hence,
() x y z
…
2/7
( )( )( ) 26 11
=
+
6
16
t 11
− 11 −
t
2
t 11
=
26
16
11
11
6
− 11 0
+t
2
− 11 1
.
26 11
+
16
z= 11
26 11
+
16 11
t and
9/9/2020
Solution
4. 4.
→ → → → Let n 1 = (2, 2, 1) and n 2 = (1, − 3, 0). Then n 1 × n 2 = (3, 1, − 8). x = 2 + 3t,
5. 5.
y = − 3 + t,
z = − 8t.
→ → Let v 1 = (1, − 2, − 1) and v 2 = ( − 1, 1, 1). Then → → v1 × v2 =
(|
−2 −1 1
1
| | , −
1
−1
−1
1
|| ,
1
−2
−1
1
|)
= ( − 1, 0, − 1).
The equation of the line is x = 2 − t, y = − 3, z = − t. 6. 6. By (6.3.1), we have a. x + 3y + 5z = (1)(1) + 3( − 1) + (5)(0) = − 2. b. − x + 0y − z = ( − 1)(1) + 0( − 1) − 3 = − 4. c. − x + y − z = ( − 1)(0) + (1)( − 1) − 2 = − 3. d. x + y − z = (1)(1) + (1)(0) − 0 = 1. 7. 7. 8. 8.
Let → n = (4, − 1, 3). Then 4x − y + 3z = (4, − 1, 3) · (2, 7, − 1) = − 2. → → a. P 1P 2 = (1, 1, 2) and P 1P 3 = (2, − 3, 3). By computation, we obtain → → n : = P 1P 2 × P 1P 3 = (9, 1, − 5).
→
By (6.3.1), we have 9x + y − 5z = (9, 1, − 5) · (1, 2, − 1) = 16. → → b. P 1P 2 = ( − 1, 1, − 2) and P 1P 3 = (0, 1, − 3). By computation, we obtain
…
3/7
9/9/2020
Solution
→ → n = P 1P 2 × P 1P 3 = ( − 1, − 3, − 1).
→
By (6.3.1), we have − x − 3y − z = ( − 1, − 3, − 1) · (1, 0, 1) = − 2.
9. 9.
→ → Let n 1 = (2, 1, 1) and n 1 = (1, − 2, 3). Then → → n1 × n2 =
(| | | | | |) 1
1
−2 3
2 1
, −
1 3
,
2 1
1 −2
= (5, − 5, − 5) = 5(1, − 1, − 1). → → Let v 1 = (1, − 1, − 1) and v 2 = (6, 0, 1). Then → → v1 × v2 =
(|
−1 −1 0
1
| | , −
1 −1 6
1
| | |) ,
1 −1 6
0
= ( − 1, − 7, 6).
Hence, − x − 7y + 6z = ( − 1, − 7, 6) · (3, 4, 0) = − 31. 10. 10. Solving the following system
{ we obtain x =
5 2
−
13 4
t, y = −
5 2
1
(
4/7
2x + 2y + 7z = 0,
− 4 t and z = t. Let t = 2. Then (x, y, z) = ( − 4, − 3, 2) is a point on the
above line, denoted by Q. Let → v = ( − 4) −
…
x − y + 3z = 5
13 4
1
)
, − 4 , 1 = (13, 1, − 4). Then → v is parallel to the above
9/9/2020
Solution
line, so it is parallel to the plane we seek. Because P and Q are in the plane we seek, so → PQ = ( − 4, − 3, 2) − (1, 4, 4) = ( − 5, − 7, − 2) is parallel to the plane we seek. Therefore, −4
−7 −2
| | , −
13 − 4 −5 −2
|| ,
13 − 1 −5 −7
|)
= − 2(15, − 23, 43).
n
(|
1
Let → n = (15, − 23, 43). Then
→ → Let n 1 = (1, 1, 1) and n 2 = ( − 1, 2, 3) be the normal vectors of the two given planes. By
tri
11. 11.
bu
15x − 23y + 43z = (15, − 23, 43) · (1, 4, 4) = 95.
(| | | 1 1 2 3
, −
1
1
−1 2
| | |) ,
1
1
−1 2
= (1, − 4, 3).
rD
→ → n1 × n2 =
is
computation, we have
ot
fo
The equation of the plane is x − 4y + 3z = (1, − 4, 3) · (1, 2, − 1) = − 10. 12. 12. Let → n = ( − 3, 1, 2). Then − 3x + y + 2z = ( − 3, 1, 2) · (1, 1, 3) = 4. → → 13. 13. Let v 1 = (1, 0, 2) and v 1 = (1, 1, 1). Then
N
→ → v 1 × v 2 = (1, 0, 2) × (1, 1, 1) = ( − 2, 1, 1) and − 2x + y + z = ( − 2, 1, 1) · (3, 5, 5) = 4. 14. 14. a. D = …
5/7
tio
→ v × PQ =
→
| (1) (1) + ( −1) ( −2) + (2) ( −1) +1|
√
12 + ( − 1 ) 2 + 22
=
√6 3
.
9/9/2020
Solution
b. D = c. D = d. D =
| (2) (0) + (1) (1) + (3) (2) −4|
√2 2 + 1 2 + 1 2
=
| (2) (1) + ( −5) (0) + (1) (1) −3|
√
22 + ( − 5 ) 2 + 12
3√14 14
= 0.
| ( −1) ( −1) + (2) ( −1) + (3) (1) −1|
√
.
√14
=
( − 1 ) 2 + 22 + 32
14
.
15. 15. 1
a. In the second equation, let y = z = 0. Then we have x = 3 . So P 0 second plane. The distance between the point P 0
(
1 3
(
1 3
)
, 0, 0 is a point in the
)
, 0, 0 and the first plane equals the
distance between the two parallel planes. By (6.3.6), we have
| () (1)
D=
1
3
+ ( − 1)(0) + (2)(0) + 3
√1
2
2
+ ( − 1) + 2
|
=
2
5√6 9
.
(
5
5
)
b. In the second equation, let y = z = 0. Then we have x = − 2 . So P 0 − 2 , 0, 0 is a point in the second plane. By (6.3.6), we have
| ( ) 5
(1) − 2
D=
…
6/7
√
+ (1)(0) + ( − 1)(0) − 1
1 2 + 1 2 + ( − 1) 2
|
=
7√3 6
.
9/9/2020
…
7/7
Solution
9/9/2020
A.7 Bases and dimensions
A.7 Bases and dimensions Section 7.1 = (2, 4, 6)
T
→ T , a2 = (0, 0, 0) ,
→
and a3
= (1, 2, 3)
T
.
→
is linearly dependent. →
→
→
2. Determine whether {a1 , a2 , a3 , a4 } is linearly dependent, where ⎛
⎞
⎛
0
⎞
⎝
⎝
⎝
0
⎠
3 −4
⎞ ⎟, ⎠
→ → → = {a1 , a2 , a3 }
7 ⎛ ⎞ → a2 = ⎜ 9 ⎟ , ⎝
⎝
−1 ⎛ ⎞ → a3 = ⎜ −2 ⎟ ,
⎝
0 −1
1
⎠
8
⎠
⎝
1
6 ⎛ ⎞ → a4 = ⎜ −2 ⎟ .
⎠
⎝
is linearly independent, where −1
⎞
→ ⎜2⎟ a4 = ⎜ ⎟. ⎜0⎟
⎠
0
0
is linearly independent, where
ot
1
⎠
fo
→ → → → = {a1 , a2 , a3 , a4 }
0
⎛
rD is tri
→ ⎜2⎟ a3 = ⎜ ⎟, ⎜0⎟
⎛ → a1 = ⎜
1/2
1
→ ⎜0⎟ a2 = ⎜ ⎟, ⎜1⎟
⎝
…
⎛
→ ⎜2⎟ a1 = ⎜ ⎟, ⎜1⎟
⎛ → a1 = ⎜
4. Determine that S
⎞
N
3. Determine whether S
1
bu t
→
⎞ ⎟, ⎠
3 ⎛ ⎞ → a2 = ⎜ −1 ⎟ , ⎝
−6
⎠
→
→
Determine whether {a1 , a2 , a3 }
io n
→
1. Let a1
1 ⎛ ⎞ → a3 = ⎜ 2 ⎟ . ⎝
3
⎠
5
⎠
9/9/2020
…
2/2
A.7 Bases and dimensions
9/9/2020
Solution
Solution 1. 1.
→
Because { a
1,
→ → a 2, a 3}
contains a zero vector
→ → → → a 2 , { a 1 , a 2 , a 1 }
is linearly
Because
→ → → → → → a 3 = a 1 − a 2 , { a 1 , a 2 , a 3 }
Because n = 4 > 3 = m, by Theorem 7.1.1, S is linearly independent. Because ∣ −1 |A| =
∣
1∣
3
∣ −1
−6
∣ −1
3
1
−1
2
0
−11 ∣
∣
0
0
∣0
−1
∣ ∣ ∣
3
−1 −9
1∣
∣ 2 R2 (−9) + R3 ∣––––––––––––– ––––––––––––– 7∣
= −11 ≠ 0,
Section 7.2 = (1, − 1, 1, 0),
→
and a2
N
ot
by Theorem 7.1.5, S is linearly independent.
→
∣ −1
∣ ∣ 2 R1 (−1) + R3 0 ∣––––––––––––– ∣ ––––––––––––– ∣0 8∣
∣
∣
1. Let a1
rD is tri
3. 3. 4. 4.
is linearly dependent.
bu t
→ → → → { a 1, a 2, a 3, a 4}
is linearly dependent. By Corollary 7.1.2,
fo
2. 2.
io n
dependent.
= (1, 0, 1, 1).
Determine whether S
→ → = {a1 , a2 }
span S. If so, find dim(span S). 2. Determine whether S
…
1/2
→ → → = {a1 , a2 , a3 }
is a basis of span S, where
i.
→ → a1 = (1, 0, − 1, 1), a2 = (0, 0, 1, 1)
ii.
→ → a1 = (1, 0, 1, 1), a2 = (0, 0, 1, − 1)
→
and a3 and
= (−2, 1, 0, 1)
→ a3 = (1, 0, 2, 0)
.
.
is a basis of
9/9/2020
Solution →
3. Let a1
→ → = (1, − 1), a2 = (0, 1), a3 = (−3, 2)
→ → → → S = {a1 , a2 , a3 , a4 } →
4. Let a1
= (1, 0, 1)
= (1, 0).
Determine whether
is a basis of span S.
→
and a2
= (1, 1, 1).
Find a vector S
to prove Theorem 7.2.2 to prove Theorem 7.2.4
→ → → = {a1 , a2 , b }
is a basis of span S.
. .
N
ot
fo
rD
is tri bu tio
n
5. Use Theorem 7.1.1 6. Use Theorem 7.1.1
→
and a4
…
2/2
9/9/2020
Solution
Solution → 1. 1. Because a 1 is not parallel to a 2, S is linearly independent. Hence, S is basis of span S and dim(span S) = 2. →→→ 2. 2. Let A = (a 1 a 2 a 3). Then
R2 , 3
−2 1 0
1
1 0 −1 1 0 1 −2 3 0 0 1
1
) ( ) →
0 0 1
)
bu
1
1 0 −1 1
1
0 1 −2 3
tri
0 1
fo
→
0
R1 ( − 2 ) + R3
is
( (
0 −1 1
rD
AT =
1
tio
n
→
N
ot
and r(A T) = 3. By Theorem 2.5.5, r(A) = r(A T) = 3. It follows from Theorem 7.2.1 that S is not a basis of span S and by Definition 7.2.1, dim(span S) = 3. 3. 3. →→→ i. Let A = (a 1 a 2 a 3). Then
…
1/6
9/9/2020
Solution
(
) ( ( )
0 0 1 −1
R1 ( − 1 ) + R3
→
1 0 2 0
R1 ( − 1 ) + R3
→
1 0 1 1 0 0 1 −1 0 0 1 −1
)
1 0 1 1
0 0 1 −1
n
AT =
1 0 1 1
tio
0 0 0 0
→
rD
( ) ( 1 1 b1
(A | b) =
is
→→ → → →T → Let A = (a 1 a 2) and b = (b 1, b 2, b 3) . Then
0 1 b2
R1 ( − 1 ) + R3
→
→
0 1
b2
0 0 b3 − b1
N
|
b1
ot
1 1 b3
1 1
fo
4. 4.
tri
bu
and r(A T) = 2. By Theorem 2.5.5, r(A) = r(A T) = 2 < 3 = n. It follows from Theorem 7.2.1 that S is not a basis of span S. ii. Because n = 4 > 2 = m, by Theorem 7.2.1, S is not a basis of span S.
|
)
= (B | → c).
→
If b 3 − b 1 ≠ 0, and r(A) = 2 and r(A b) = 3. Hence, r(A) < r(A b). By Theorem 4.5.1 (3), we see that if b 1 ≠ b 3 and b 2 is a real number, then the system has no solutions. Hence we can choose b 1 = b 2 = 1 →
T
and b 3 = 0, that is, b = (1, 1, 0) By Theorem 7.2.4, S =
…
2/6
{ } → → → a 1, a 2 b
is a basis of span S.
9/9/2020
Solution
5. 5.
Let S =
{
→ → → a 1, a 2⋯, a n
}
→
be a basis of span S. We will prove that for each b ∈ span S, there exists
a unique vector (x 1, x 2⋯, x n) such that (A.7.1) → → → b = x 1a 1 + x 2a 2 + ⋯x na n.
→
→
Because b ∈ span S, there exists a vector (x 1, x 2⋯x n) such that (A.7.1) holds. Assume that there exists another vector (y 1, y 2, ⋯, y n) such that (A.7.2) → → → b = y 1a 1 + y 2a 2 + ⋯y na n.
→
Then → → → → → → x 1a 1 + x 2a 2 + ⋯x na n = y 1a 1 + y 2a 2 + ⋯y na n and (A.7.3) → → → → (x 1 − y 1)a 1 + (x 2 − y 2)a 2 + ⋯(x n − y n)a n = 0. Because S is a basis of span S, S is linearly independent. By (A.7.3) and Theorem 7.1.1, we obtain x i − y i = 0 for i ∈ I n. This implies that x i = y i for i ∈ I n. Hence, the vectors satisfying (A.7.1) are unique. …
3/6
9/9/2020
Solution
6. 6.
Let S =
{
→ → → a 1, a 2, ⋯, a n
}
→
be a basis of span S. Assume that S ≠ ℝ m. Let b ∈ span S. We prove
that
S1 =
{
→ → → → a 1, a 2, ⋯ a n, b
}
is linearly independent. Indeed, assume that there exists a vector (x 1, x 2, ⋯, x n, x n + 1) such that (A.7.4) → → → → → x 1a 1 + x 2a 2 + ⋯x na n + x n + 1 b = 0. →
→
Because b ∈ span S, we have x n + 1 = 0. If not, then x n + 1 ≠ 0 and by (A.7.4), b is a linear combination →
of S. Hence, we obtain b ∈ span S, a contradiction. Because x n + 1 = 0, it follows from (A.7.4) that (A.7.5) → → → → x 1a 1 + x 2a 2 + ⋯x na n = 0. Because S is a basis of span S, S is linearly independent and by (A.7.5), x i = 0 for i ∈ I n. Hence, (A.7.4) only has the zero solution. By Theorem 7.1.1, S is linearly independent.
Section 7.3 1. Find a basis for the column space of C, where
…
4/6
9/9/2020
Solution
(
4
0 0 1 3 −2 −6 0 0 0 0
1
5
0 0 0 0
0
0
)
.
n
C=
1 −3 4 −2 5
bu
2 4 1 −1 0
1 −3 4 −2 5 2 −6 9 −1 8 2 −6 9 −1 9
4 2 7
fo
(
( ) ) 0 0 0 3 1
.
−1 3 −4 2 −5 −4
ot
A3 =
1
rD
2 0
A2 =
1 1 1 0 0
is
A1 =
1 −1 3
tri
( )
1 3 0 2 0
0 2 −1
tio
2. For each of the following matrices, find a set of its column vectors that forms a basis for its column space.
N
3. For each of the following row echelon matrices, find the basis and dimension of its row space and use the basis vectors to denote the row space.
…
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9/9/2020
Solution
D1 =
0 1 0 0 0 0
D2 =
( ) ( ) 0 0 0 1 3 0 0 0 0 0
1 2 5 0 3 0 1 3 0 0
D3 =
0 0 0 1 0
0 0 0 0 0
0 0 0 0 0
n
( )
0 1 −2 0 1
1 1 0
bu tio
4. Find a set of basis vectors for R B and dimR B, where
( ) 1 −1 3
{
}
0 4 . −1 −3 1
is rD
5. Let S =
→ → → → → a 1, a 2, a 3, a 4, a 5 , where
2
tri
B=
fo
→ → → T T a 1 = (1, − 2, 0, 3) , a 2 = (2, − 5, − 3, 6) , a 3 = (0, 1, 3, 0) T,
ot
→ → T a 4 = (2, − 1, 4, − 7) , a 5 = (5, − 8, 1, 2) T.
N
a. Find a subset of S that forms a basis for the spanning space span S. b. Find another basis for span S.
…
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9/9/2020
Solution
Solution → 1. 1. Because a 1 is not parallel to a 2, S is linearly independent. Hence, S is basis of span S and dim(span S) = 2. →→→ 2. 2. Let A = (a 1 a 2 a 3). Then
R2 , 3
−2 1 0
1
1 0 −1 1 0 1 −2 3 0 0 1
1
) ( ) →
0 0 1
)
bu
1
1 0 −1 1
1
0 1 −2 3
tri
0 1
fo
→
0
R1 ( − 2 ) + R3
is
( (
0 −1 1
rD
AT =
1
tio
n
→
N
ot
and r(A T) = 3. By Theorem 2.5.5, r(A) = r(A T) = 3. It follows from Theorem 7.2.1 that S is not a basis of span S and by Definition 7.2.1, dim(span S) = 3. 3. 3. →→→ i. Let A = (a 1 a 2 a 3). Then
…
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9/9/2020
Solution
(
) ( ( )
0 0 1 −1
R1 ( − 1 ) + R3
→
1 0 2 0
R1 ( − 1 ) + R3
→
1 0 1 1 0 0 1 −1 0 0 1 −1
)
1 0 1 1
0 0 1 −1
n
AT =
1 0 1 1
tio
0 0 0 0
→
rD
( ) ( 1 1 b1
(A | b) =
is
→→ → → →T → Let A = (a 1 a 2) and b = (b 1, b 2, b 3) . Then
0 1 b2
R1 ( − 1 ) + R3
→
→
0 1
b2
0 0 b3 − b1
N
|
b1
ot
1 1 b3
1 1
fo
4. 4.
tri
bu
and r(A T) = 2. By Theorem 2.5.5, r(A) = r(A T) = 2 < 3 = n. It follows from Theorem 7.2.1 that S is not a basis of span S. ii. Because n = 4 > 2 = m, by Theorem 7.2.1, S is not a basis of span S.
|
)
= (B | → c).
→
If b 3 − b 1 ≠ 0, and r(A) = 2 and r(A b) = 3. Hence, r(A) < r(A b). By Theorem 4.5.1 (3), we see that if b 1 ≠ b 3 and b 2 is a real number, then the system has no solutions. Hence we can choose b 1 = b 2 = 1 →
T
and b 3 = 0, that is, b = (1, 1, 0) By Theorem 7.2.4, S =
…
2/6
{ } → → → a 1, a 2 b
is a basis of span S.
9/9/2020
Solution
5. 5.
Let S =
{
→ → → a 1, a 2⋯, a n
}
→
be a basis of span S. We will prove that for each b ∈ span S, there exists
a unique vector (x 1, x 2⋯, x n) such that
→ → → b = x 1a 1 + x 2a 2 + ⋯x na n.
n
(A.7.1)
→
rib ut io
→
Because b ∈ span S, there exists a vector (x 1, x 2⋯x n) such that (A.7.1) holds. Assume that there exists another vector (y 1, y 2, ⋯, y n) such that (A.7.2)
is t
→ → → b = y 1a 1 + y 2a 2 + ⋯y na n.
rD
→
Then
N
and (A.7.3)
ot
fo
→ → → → → → x 1a 1 + x 2a 2 + ⋯x na n = y 1a 1 + y 2a 2 + ⋯y na n
→ → → → (x 1 − y 1)a 1 + (x 2 − y 2)a 2 + ⋯(x n − y n)a n = 0. Because S is a basis of span S, S is linearly independent. By (A.7.3) and Theorem 7.1.1, we obtain x − y i = 0 for i ∈ I n. This implies that x i = y i for i ∈ I n. Hence, the vectors satisfying (A.7.1) are unique. …
3/6
9/9/2020
Solution
6. 6.
Let S =
{
→ → → a 1, a 2, ⋯, a n
}
→
be a basis of span S. Assume that S ≠ ℝ m. Let b ∈ span S. We prove
that
}
n
{
rib ut io
S1 =
→ → → → a 1, a 2, ⋯ a n, b
is linearly independent. Indeed, assume that there exists a vector (x 1, x 2, ⋯, x n, x n + 1) such that (A.7.4)
is t
→ → → → → x 1a 1 + x 2a 2 + ⋯x na n + x n + 1 b = 0. →
→
rD
Because b ∈ span S, we have x n + 1 = 0. If not, then x n + 1 ≠ 0 and by (A.7.4), b is a linear combination →
of S. Hence, we obtain b ∈ span S, a contradiction. Because x n + 1 = 0, it follows from (A.7.4) that
fo
(A.7.5)
ot
→ → → → x 1a 1 + x 2a 2 + ⋯x na n = 0.
N
Because S is a basis of span S, S is linearly independent and by (A.7.5), x i = 0 for i ∈ I n. Hence, (A.7.4) only has the zero solution. By Theorem 7.1.1, S is linearly independent.
Section 7.3 1. Find a basis for the column space of C, where
…
4/6
Solution
C=
(
1 −3 4 −2 5
4
0 0 1 3 −2 −6 0 0 0 0
1
5
0 0 0 0
0
0
)
.
ut io n
9/9/2020
2. For each of the following matrices, find a set of its column vectors that forms a basis for its column space.
(
( ) )
rib
2 4 1 −1 0
0 0 0 3 1
1 −3 4 −2 5 2 −6 9 −1 8
2
2 −6 9 −1 9
4 7
.
−1 3 −4 2 −5 −4
N ot
A3 =
1
D
2 0
A2 =
1 1 1 0 0
fo r
A1 =
1 −1 3
is t
( )
1 3 0 2 0
0 2 −1
3. For each of the following row echelon matrices, find the basis and dimension of its row space and use the basis vectors to denote the row space.
…
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9/9/2020
Solution
D1 =
0 1 0 0 0 0
D2 =
( ) ( ) 0 0 0 1 3 0 0 0 0 0
1 2 5 0 3
D3 =
0 0 0 0 0
0 1 3 0 0 0 0 0 1 0 0 0 0 0 0
n
( )
0 1 −2 0 1
1 1 0
( )
B=
5. Let S =
{
}
→ → → → → a 1, a 2, a 3, a 4, a 5 , where
2
rD is tri bu t
1 −1 3
io
4. Find a set of basis vectors for R B and dimR B, where
0 4 . −1 −3 1
fo
→ → → T T a 1 = (1, − 2, 0, 3) , a 2 = (2, − 5, − 3, 6) , a 3 = (0, 1, 3, 0) T,
ot
→ → T a 4 = (2, − 1, 4, − 7) , a 5 = (5, − 8, 1, 2) T.
N
a. Find a subset of S that forms a basis for the spanning space span S. b. Find another basis for span S.
…
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9/9/2020
Solution
Solution 1. 1.
→ We denote the column vectors of C by C i , i = 1, 2, 3, 4, 5, 6. Then →→→→→→ B = (C 1C 2C 3C 4C 5C 6).
→ → → The column vectors containing the leading entries are C 1, C 3, C 5. Hence,
{
→ → → C 1, C 3, C 5
}
forms a basis
for the column space of C. 2. 2.
( )( ) ( ) 0
A1 =
2
−1
1 −1
3
2
1
0
R1 ( − 1 ) + R3
→
R1 , 2
→
1 −1 0
2
−1
2
0
1
()
( )( 1 −1
R1 ( − 2 ) + R3
3
3
0
2
−1
0
0
−4
R3
1
2
→
R3
−
1 4
)
1 −1
→ →
0
2
−1
0
2
−5
( ) 1 −1
3
3
1
0
1
−2
0
0
1
*
= C1 .
→ → → Because the column vectors C 1, C 2, C 3 of C 1* contain the leading entries, by Theorem 7.3.1, the pivot → → → column vectors 1 2 a 3 of A 1 form a basis for C A . 1
…
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9/9/2020
Solution
) ( ( ) ( ) (
A2 =
(
1 3 0 2
0
1 1 1 0
0
2 4 1 −1 0 0 0 0 3
1
1 3
R1 ( − 1 ) + R3
→
1 3
R1 ( − 1 ) + R2
→
R1 ( − 2 ) + R3
0 2
0 0
0 −3 0
0 0
0 3
0
0 −2 1 −2 0 0 0
0 −3 0
0 0
0 0
1
R2
0
0 −2 1 −2 0 0 −2 1 −5 0 0 0
0 3
1
)
R1 ( 1 ) + R4
→
1
( ) 1
−2
→
R3
0 2
0
0 −2 1 −2 0
0 2
1 3
( ) 1
−3
1 3 0
2 0 1
0 1 −2 1 0
0 0 0
1 0
0 0 0
0 1
)
= C 2* .
→ → → → * Because the column vectors C 1, C 2, C 4, C 5 of C 2 contain the leading entries, by Theorem 7.3.1, the → → → → pivot column vectors a 1, a 2, a 4, a 5, of A 2 form a basis for C A . 2
→ We denote the column vectors of A 3 by a i , i = 1, 2, 3, 4, 5, 6. Then →→→→→→ A 3 = (a 1 a 2 a 3 a 4 a 5 a 6).
…
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Solution
*
By using row operations, we can change A 3 into a row echelon matrix C 3 , which is the same as → → → * Exercise 1 above. Hence, the column vectors C 1, C 3, C 5 of C 3 contain leading entries. → → → By Theorem 7.3.1, the pivot column vectors a 1, a 3, a 5 of A 3 form a basis of C A . 3
3. 3. RD
The basis vectors for R D
1
1
→ → are d 1 = (1, 1, 0) and d 2 = (0, 1, 0). Hence, dim(R D ) = 2 and 1
{ }
→ → = span d 1, d 2 .
→ → The basis vectors for R D 2 are d 1 = (0, 1, − 2, 0, 1) and d 2 = (0, 0, 0, 1, 3). Hence, dim(R D ) = 2 and 2
RD
1
{ }
→ → = span d 1, d 2 . → d 1 = (1, 2, 5, 0, 3),
Hence, dim(R D ) = 3 and R D 3
4. 4.
…
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3
→ → d 2 = (0, 1, 3, 0, 0), d 3 = (0, 0, 0, 1, 0).
→ → → = span{d 1, d 2, d 3}.
9/9/2020
Solution
( ) ( ) () ( ) ( )
B =
1
−1 3
2
0
R1 ( − 2 ) + R2
4
→
R1 ( 1 ) + R3
−1 −3 1
1 −1 3 0 2
−2
0 0
0
1
R2
1 −1 3 0 2
R2 ( 2 ) + R3
−2
→
0 −4 4
1 −1 3
2
→
0 1
−1
0 0
0
= D.
{ }
→ → → → → → Let d 1 = (1, − 1, 3), d 2 = (0, 1, − 1). Then d 1, d 2 are basis vectors for ℝ B. ℝ B = span d 1, d 2 dim(R B) = 2. 5. 5. a. To find a subset of S that forms a basis for span S, we need to put these vectors → → → → → a 1, a 2, a 3, a 4, a 5 as the column vectors of a matrix A, that is,
A=
(
1
2
0
2
5
−2 −5 1 −1 −8 0 3
−3 3 6
4
0 −7
1 2
)
.
We use row operations to change A into a row echelon matrix.
…
4/9
and
9/9/2020
Solution
) ( ) ( ) ( ) ( ) 5
1 2
R1 ( 2 ) + R2
0
−3 3 4
3
6
1
→
0 2
5
0 −1 1 3
2
0 0
0 −5
−5
0 0
0 − 13 13
0 2
5
0 −1 1 3
2
0 0
0 1
1
0 0
0 − 13 13
R3
R3
( ) 1
− 13
→
0 −1 1 3
2
0 −3 3 4
1
0 0
0 − 13 13
( ) 1
−5
→
+ R4
1 2
0 2 5
0 −1 1 3 2 0 0
0 1 1
0 0
0 0 0
= C.
fo r
1 2
R1 ( − 3 ) + R4
0 −7 2
1 2
R1 ( − 3 ) + R3
→
5
ut io n
−2 −5 1 −1 −8
0 2
rib
0 2
is t
2
D
(
A =
1
→ → → The column vectors C 1, C 2, C 4 of C contain the leading entries of C. By Theorem 7.3.1, the
{
}
N ot
pivot column vectors
→ → → a 1, a 2, a 4
of A form a basis of span S.
b. We can use Theorem 7.3.2 to find a basis for span S. We need to put these vectors → → → → → a 1, a 2, a 3, a 4, a 5 as row vectors of a matrix B, that is,
…
5/9
9/9/2020
Solution
( ) 1 −2
B=
0
3
2 −5 −3
6
0
3
0
2 −1
4
−7
5 −8
1
2
1
.
We use row operations to change B into a row echelon matrix D.
( ) ( )( ) ( ) ( ) 1 −2 0
R1 ( − 2 ) + R2
B
0 −1 −3 0
→
0 1
3
0
R1 ( − 2 ) + R4
0 3
4
− 13
0 2
1
− 13
R1 ( − 5 ) + R5
1 −2 0
3
0 −1 −3 0 0 0
0
0
0 3
4
− 13
0 2
1
− 13
1 −2 0
6/9
0 0
− 5 − 13
0 0
5
− 13
R2 ( 1 ) + R3
→
1 −2 0
R3 , 4
→
R4 , 5
3
0 −1 −3 0 0 3
4
− 13
0 2
1
− 13
0 0
0
0
3
0 −1 −3 0
…
3
1 −2 0
R3 ( 1 ) + R4
→
R2 ( − 3 ) + R3
→
R2 ( − 2 ) + R4
3
0 −1 −3 0 0 0
− 5 − 13
0 0
0
0
( ) ( )
9/9/2020
Solution
0 0
0
0
0 0
( )
0
0
1 −2 0 3
R2 ( − 1 )
→ R3
( ) 1
−5
0 1
3 0
0 0
1
0 0
0 0
0 0
0 0
13 5
= D.
(
→ → → Let d 1 = (1, − 2, 0, 3), d 2 = (0, 1, 3, 0), and d 3 = 0, 0, 1,
{
→ → → d 1, d 2, d 3
}
13 5
)
By Theorem 7.3.2,
forms a basis for span S.
Section 7.4 1. For each of the following pairs of vectors, verify that it is a basis of ℝ 2 and find the coordinates of the →
vector b = (x, y) T.
S1 =
S3 =
…
7/9
{( ) ( )} {( ) ( )} {( ) ( )} {( ) ( )} −1 1
5
−6
,
,
1 0
4
−1
S2 =
S4 =
−3 2
2
,
3
−2
3
,
−2 5
9/9/2020
Solution
→ → → 2. Let a 1 = (1, 0, 1) T, a 2 = (0, 1, 1) T, and a 3 = (1, 1, 0) T. i. Show that S =
{
→ → → a 1, a 2, a 3
→
}
is a basis of ℝ 3.
→
ii. Let b = (1, 1, 1) T. Find ( b) S. → → → 3. Let a 1 = (1, 0, 0) T, a 2 = (2, 4, 0) T, and a 3 = (3, − 1, 5) T. i. Show that S =
{
→ → → a 1, a 2, a 3
}
→
is a basis of ℝ 3. →
ii. Let b = (2, − 1, − 2). Find ( b) S.
4. Let S 1 =
{( ) ( )} −1 1
,
1 0
and S 2 =
→
{( ) ( )} −3 2
→
i. If ( b) S = (2, − 1), find ( b) S . 2
1
→
→
ii. If ( b) S = (13, − 26), find ( b) S . 1
→
2
→
5. If ( b) T = (1, 2, 3), find ( b) S, where
…
8/9
,
2 3
.
Solution
{( ) ( ) ( )} {( ) ( ) ( )}
T=
i. Show that S : = →
{
0
0
0 , 1
1
0
1
0 , 0
1
)
→ , and b 3 =
3 5
→ → → b 1, b 2, b 3
}
(
,
1
1
,
−1
3
.
1
4
, 0, 5
,
5
)
.
is an orthonormal basis of ℝ 3.
is t
(
0
1
S=
→ → 4 6. Let b 1 = (0, 1, 0), b 2 = − 5 , 0,
1
n
−1
rib ut io
9/9/2020
→
rD
ii. Let b = (1, − 1, 1). Find ( b) S.
{
1
N
→
( b) S .
2. Find an orthonormal basis S 2 : =
…
9/9
→ → → b 1, b 2, b 3
}
of ℝ 3 by using the Gram-Schmidt process and
ot
1. Find an orthogonal basis S 1 : =
fo
→ → → → 7. Let a 1 = (1, 1, 1), a 2 = (0, 1, 1), a 3 = (0, 0, 1), and b = ( − 1, 1, − 1) T.
{
→ → → q 1, q 2, q 3
}
→
of ℝ 3 by normalizing S 1 and ( b) S . 2
9/9/2020
Solution
Solution 1. 1.
→ We denote the column vectors of C by C i , i = 1, 2, 3, 4, 5, 6. Then
tri bu tio
n
→→→→→→ B = (C 1C 2C 3C 4C 5C 6).
→ → → The column vectors containing the leading entries are C 1, C 3, C 5. Hence,
}
forms a basis
( )( ) ( ) 1 −1
3
2
1
0
→
R1 , 2
→
0
2
2
0
0 0
R1 ( − 2 ) + R3
3
1 −1
→ →
−1 1
()
( )( 1 −1
3
2
−1
0
−4
N
R1 ( − 1 ) + R3
1 −1
rD
−1
fo
2
R3
1
ot
0
A1 =
is
for the column space of C. 2. 2.
{
→ → → C 1, C 3, C 5
2
→
R3
−
1 4
)
0
2
−1
0
2
−5
( ) 1 −1
3
3
1
0
1
−2
0
0
1
*
= C1 .
→ → → Because the column vectors C 1, C 2, C 3 of C 1* contain the leading entries, by Theorem 7.3.1, the pivot → → → column vectors a 1, a 2, a 3 of A 1 form a basis for C A . 1
…
1/9
Solution
0 0 0 3
1
1 3
R1 ( − 1 ) + R3
→
1 3
→
R1 ( − 2 ) + R3
0 2
0 0
0 −3 0
0 0
0 3
0
0 −2 1 −2 0 0 0
0 −3 0
0 0
0 0
0 3
1
)
R1 ( 1 ) + R4
→
1
( ) 1
−2
→
R3
0 0
( ) 1
−3
1 3 0
2 0 1
0 1 −2 1 0
0 0 0
1 0
0 0 0
0 1
fo
1
R2
0 −2 1 −5 0
0
0 −2 1 −2 0
0 2
0 −2 1 −2 0
tio
2 4 1 −1 0
R1 ( − 1 ) + R2
0
bu
0
0 2
)
tri
1 1 1 0
1 3
rD
(
0
n
) ( ( ) ( ) (
A2 =
1 3 0 2
is
9/9/2020
= C 2* .
N
ot
→ → → → * Because the column vectors C 1, C 2, C 4, C 5 of C 2 contain the leading entries, by Theorem 7.3.1, the → → → → pivot column vectors a 1, a 2, a 4, a 5, of A 2 form a basis for C A . 2
→ We denote the column vectors of A 3 by a i , i = 1, 2, 3, 4, 5, 6. Then →→→→→→ A 3 = (a 1 a 2 a 3 a 4 a 5 a 6).
…
2/9
9/9/2020
Solution
*
By using row operations, we can change A 3 into a row echelon matrix C 3 , which is the same as → → → * Exercise 1 above. Hence, the column vectors C 1, C 3, C 5 of C 3 contain leading entries. → → → By Theorem 7.3.1, the pivot column vectors a 1, a 3, a 5 of A 3 form a basis of C A . 3
RD
1
1
1
{ }
→ → = span d 1, d 2 .
n
The basis vectors for R D
rib ut io
3. 3.
→ → are d 1 = (1, 1, 0) and d 2 = (0, 1, 0). Hence, dim(R D ) = 2 and
→ → The basis vectors for R D 2 are d 1 = (0, 1, − 2, 0, 1) and d 2 = (0, 0, 0, 1, 3). Hence, dim(R D ) = 2 and 2
{ }
→ d 1 = (1, 2, 5, 0, 3),
N
4. 4.
3
→ → → = span{d 1, d 2, d 3}.
fo
3
→ → d 2 = (0, 1, 3, 0, 0), d 3 = (0, 0, 0, 1, 0).
ot
Hence, dim(R D ) = 3 and R D
…
3/9
is t
1
rD
RD
→ → = span d 1, d 2 .
9/9/2020
Solution
( ) ( ) () ( ) ( ) 2
0
R1 ( − 2 ) + R2
4
→
R1 ( 1 ) + R3
−1 −3 1
1 −1 3
1
R2
0 2
−2
0 0
0
1 −1 3 0 2
R2 ( 2 ) + R3
−2
→
0 −4 4
1 −1 3
2
→
0 1
−1
0 0
0
= D.
n
−1 3
rib ut io
B =
1
{ }
→ → → → → → Let d 1 = (1, − 1, 3), d 2 = (0, 1, − 1). Then d 1, d 2 are basis vectors for ℝ B. ℝ B = span d 1, d 2
is t
dim(R B) = 2. 5. 5.
fo
(
1
2
N
0
2
5
−2 −5 1 −1 −8
ot
A=
rD
a. To find a subset of S that forms a basis for span S, we need to put these vectors → → → → → a 1, a 2, a 3, a 4, a 5 as the column vectors of a matrix A, that is,
0 3
−3 3 6
4
0 −7
1 2
)
.
We use row operations to change A into a row echelon matrix.
…
4/9
and
9/9/2020
Solution
) ( ) ( ) ( ) ( ) 5
1 2
R1 ( 2 ) + R2
0
−3 3 4
3
6
1
→
0 2
5
0 −1 1 3
2
0 0
0 −5
−5
0 0
0 − 13 13
0 2
5
0 −1 1 3
2
0 0
0 1
1
0 0
0 − 13 13
R3
R3
( ) 1
− 13
→
0 −1 1 3
2
0 −3 3 4
1
0 0
0 − 13 13
( ) 1
−5
→
+ R4
1 2
0 2 5
0 −1 1 3 2 0 0
0 1 1
0 0
0 0 0
= C.
fo r
1 2
R1 ( − 3 ) + R4
0 −7 2
1 2
R1 ( − 3 ) + R3
→
5
ut io n
−2 −5 1 −1 −8
0 2
rib
0 2
is t
2
D
(
A =
1
→ → → The column vectors C 1, C 2, C 4 of C contain the leading entries of C. By Theorem 7.3.1, the
{
}
N ot
pivot column vectors
→ → → a 1, a 2, a 4
of A form a basis of span S.
b. We can use Theorem 7.3.2 to find a basis for span S. We need to put these vectors → → → → → a 1, a 2, a 3, a 4, a 5 as row vectors of a matrix B, that is,
…
5/9
9/9/2020
Solution
( ) 1 −2
B=
0
3
2 −5 −3
6
0
3
0
2 −1
4
−7
5 −8
1
2
1
.
We use row operations to change B into a row echelon matrix D.
( ) ( )( ) ( ) ( ) 1 −2 0
R1 ( − 2 ) + R2
B
0 −1 −3 0
→
0 1
3
0
R1 ( − 2 ) + R4
0 3
4
− 13
0 2
1
− 13
R1 ( − 5 ) + R5
1 −2 0
3
0 −1 −3 0 0 0
0
0
0 3
4
− 13
0 2
1
− 13
1 −2 0
6/9
0 0
− 5 − 13
0 0
5
− 13
R2 ( 1 ) + R3
→
1 −2 0
R3 , 4
→
R4 , 5
3
0 −1 −3 0 0 3
4
− 13
0 2
1
− 13
0 0
0
0
3
0 −1 −3 0
…
3
1 −2 0
R3 ( 1 ) + R4
→
R2 ( − 3 ) + R3
→
R2 ( − 2 ) + R4
3
0 −1 −3 0 0 0
− 5 − 13
0 0
0
0
( ) ( )
9/9/2020
Solution
0 0
0
0
0 0
( )
0
0
1 −2 0 3
R2 ( − 1 )
→ R3
( ) 1
−5
0 1
3 0
0 0
1
0 0
0 0
0 0
0 0
13 5
= D.
(
→ → → Let d 1 = (1, − 2, 0, 3), d 2 = (0, 1, 3, 0), and d 3 = 0, 0, 1,
{
→ → → d 1, d 2, d 3
}
13 5
)
By Theorem 7.3.2,
forms a basis for span S.
Section 7.4 1. For each of the following pairs of vectors, verify that it is a basis of ℝ 2 and find the coordinates of the →
vector b = (x, y) T.
S1 =
S3 =
…
7/9
{( ) ( )} {( ) ( )} {( ) ( )} {( ) ( )} −1 1
5
−6
,
,
1 0
4
−1
S2 =
S4 =
−3 2
2
,
3
−2
3
,
−2 5
9/9/2020
Solution
→ → → 2. Let a 1 = (1, 0, 1) T, a 2 = (0, 1, 1) T, and a 3 = (1, 1, 0) T. i. Show that S =
{
→ → → a 1, a 2, a 3
→
}
is a basis of ℝ 3.
→
ii. Let b = (1, 1, 1) T. Find ( b) S. → → → 3. Let a 1 = (1, 0, 0) T, a 2 = (2, 4, 0) T, and a 3 = (3, − 1, 5) T. i. Show that S =
{
→ → → a 1, a 2, a 3
}
→
is a basis of ℝ 3. →
ii. Let b = (2, − 1, − 2). Find ( b) S.
4. Let S 1 =
{( ) ( )} −1 1
,
1 0
and S 2 =
→
{( ) ( )} −3 2
→
i. If ( b) S = (2, − 1), find ( b) S . 2
1
→
→
ii. If ( b) S = (13, − 26), find ( b) S . 1
→
2
→
5. If ( b) T = (1, 2, 3), find ( b) S, where
…
8/9
,
2 3
.
9/9/2020
Solution
{( ) ( ) ( )} {( ) ( ) ( )}
→
{
1
0
1
0 , 0
1
)
→ , and b 3 =
3 5
→ → → b 1, b 2, b 3
}
(
1
−1
1
4
, 0, 5
.
io n
1
,
3
,
5
)
.
is an orthonormal basis of ℝ 3.
is t
i. Show that S : =
0
0
0 , 1
,
ut
T=
(
0
1
S=
→ → 4 6. Let b 1 = (0, 1, 0), b 2 = − 5 , 0,
1
rib
−1
→
rD
ii. Let b = (1, − 1, 1). Find ( b) S.
{
1
N
→
( b) S .
2. Find an orthonormal basis S 2 : =
…
9/9
→ → → b 1, b 2, b 3
}
of ℝ 3 by using the Gram-Schmidt process and
ot
1. Find an orthogonal basis S 1 : =
fo
→ → → → 7. Let a 1 = (1, 1, 1), a 2 = (0, 1, 1), a 3 = (0, 0, 1), and b = ( − 1, 1, − 1) T.
{
→ → → q 1, q 2, q 3
}
→
of ℝ 3 by normalizing S 1 and ( b) S . 2
9/9/2020
Solution
Solution 1. 1.
Let A 1 =
( ) −1 1 1
0
. Then |A 1 | = − 1 ≠ 0. It follows from Theorem (ii) that S 1 is a basis of ℝ 2. By Theorem 2.7.5,
A 1− 1 =
1
(
0
−1
−1 −1 −1
) ( ) =
0 1 1 1
.
( )( ) ( )
0 1 → − 1→ By (7.4.5), ( b) S = A 1 b = 1 1 1 Let A 2 =
( ) −3 2 2
3
x y
y
=
x+y
.
. Then |A 2 | = − 9 − 4 = − 13 ≠ 0. By Theorem 7.2.1 (ii), S 2 is a basis of ℝ 2. By Theorem 2.7.5 and
(7.4.5),
−1
A2
→
Let A 3 = (7.4.5),
…
1/9
→
(
5
4
−6 −1
)
1
(
3
−2
13 − 2 − 3
)
=
( ) 2
13
2
3
13
13
( ) ( ) 3
and ( b) S = A 2− 1 b = 2
=
3
− 13
2
− 13
13
2
3
13
13
() x y
| |=
. Then A 3
3x
=
2y
− 13 + 13 2x
13
3y
,
+ 13
− 5 + 24 = 19 ≠ 0. By Theorem 7.2.1 (ii), S 3 is a basis of ℝ 2. By Theorem 2.7.5 and
9/9/2020
Solution
−1
A3
− 1→
→
Let A 4 =
(
3
−2
−2
5
)
()
− 19 − 19 6
5
19
19
x y
=
6x
19
5y
6
5
)
=
( ) 6
5
19
19
.
,
+ 19
| | = 15 − 4 = 11 ≠ 0. By Theorem 7.2.1 (ii), S4 is a basis of ℝ2. By Theorem 2.7.5 and (7.4.5),
. Then A 4
−1
− 1→
(
4
1
− 19 − 19
4y
x
− 19 − 19
A4
→
19
−1 −4
( ) ( ) 4
1
and ( b) S = A 3 b = 3
=
1
and ( b)S 4 = A 4 b =
( ) ( ) 5
2
11
11
2
3
11
11
() x y
=
5x
2y
2x
3y
+ 11 11 11
=
1
( ) 5 2
11 2 3
=
( ) 5
2
11
11
2
3
11
11
.
.
+ 11
2. 2.
| | 1 0 1
i. Because |A | =
0 1 1 1 1 0
ii. …
2/9
= − 2 ≠ 0, by Theorem 7.2.1 (ii),
{
→ → → a 1, a 2, a 3
}
is a basis of ℝ 3.
9/9/2020
Solution
| ( →
(A 0) =
R2 ( − 1 ) + R3
→
) (
1 0 1 1 R ( −1) +R 1 0 1 3 0 1 1 1 0 1 → 1 1 0 1
(
1 0
1
1
0 1
1
1
)
R3
0 0 −2 −1
1
2
→
1
1
1
0 1 −1 0
( ) −
1
)
( ) 1 0 1 1 0 1 1 1 1
0 0 1 2
( ) 1
1 0 0 2
R2 ( − 1 ) + R2
→ R3 ( − 1 ) + R1
1
0 1 0 2
.
1
0 0 1 2
→
Hence, ( b) S =
(
1 1 1
, ,
2 2 2
)
.
3. 3.
| | { } | | ||| | 1 2
i. Because |A | =
3
0 4 −1 0 0
2
2
= 20 ≠ 0,
5
3
ii. Because |A | = − 1 4 − 1 = 78, −2 0 5 we have
…
3/9
→ → → a 1, a 2, a 3
1
A2
=
2
is a basis of ℝ 3.
3
0 −1 −1 0 −2
5
| | 1 2
= − 7 and |A 3 | =
2
0 4 −1 0 0 −2
= − 8, by Cramer’s rule
9/9/2020
Solution
|A1 |
x=
→
Hence, ( b) S = (x, y, z) =
(
=
|A|
39
7
78 20 2
39
=
10
|A2 |
, y=
|A|
= −
7 20
|A3 |
and z =
=
|A|
−8 20
2
= −
5
.
)
, − 20 , − 5 . 10
4. 4. i. Let A =
( ) −1 1 1
0
and B =
( )
(
−3 2
) ( )
0 −1 0 1 1 . Then by Theorem 2.7.5, A − 1 = − 1 = . Hence, 3 −1 −1 1 1
2
( )( ) ( ) 0 1
C = A − 1B =
1 1
−3 2 2
3
=
2
3
−1 5
( )( ) ( )
2 3 → → and ( b) S = C( b) T = −1 5
ii. Let A =
( ) −1 1 1
0
and B =
2
−1
=
1
−7
.
( )
(
−3 2 2
( )( ) ( ) 3
C = B − 1A =
By Theorem 7.4.3, we have
…
4/9
)
3 −2 1 . Then by Theorem 2.7.5, B − 1 = − 13 = 3 −2 −3
2
− 13
13
2
3
13
13
5
−1 1 1
0
=
3
13
− 13
1
2
13
13
.
( ) 3
2
− 13
13
2
3
13
13
. Hence,
9/9/2020
Solution
( )( 5
→
→
( b) S = C( b) T =
Let A =
1
0 0
0
1 1
1
0
R1 ( − 1 )
→
0
1
13
13
0
1
0
1
0
1
1
1
0
0
1
1
−1 −1 1
− 26
) ( ) 11
−3
1 1 0 0 1
0
1 1 0 0 1
1 −1 0 −1 0 0 R ( −1) +R 1 0 0 2 3 0 1 0 1 1 0 0 1 0 → 1
=
0 −1 1
1
0
0 1
0
1
1
1
1
2
R2 ( 1 ) + R1
0
1
1
1
.
−1 −2 −1
→
→
( b) S = C( b) T =
5/9
(
0
1
1
1
1
2
−1 −2 −1
) 0
)
0 . 0 0 1 −1 −1 1
By Theorem 7.4.3, we have
…
.
−1 1 0 1 0 0 R ( −1) +R −1 1 0 1 0 0 1 2 1 0 0 0 1 0 0 1 0 1 1 0 →
0
B=
2
=
1
0
C=A
1
13
1 . 0 −1 1
and B =
(A I) =
−1
− 13
( ) ( ) | ( ) ( ( ) ( ( )( ) ( ) −1 1 0
5. 5.
3
13
)( ) ( ) 1 2 3
5
=
9
−8
.
9/9/2020
Solution
| | 4
0 −5
6. 6.
Because |A | =
1 0
3
5
0
0
3
4
5
5
→ → → → → → → → → = 1 ≠ 0, b 1 ⋅ b 2 = 0, b 1 ⋅ b 3 = 0, b 2 ⋅ b 3 = 0, ∥ b 1 ∥ = 1, ∥ b 2 ∥ = 1, ∥ b 3 ∥ = 1, S is
an orthonormal basis of ℝ 3. Because
→
x =
→
b ⋅ b1
∥
() 0
= (1, − 1, 1) 1 0 b1 ∥ 2 →
= − 1,
() 4
→
y =
b ⋅ b2
∥
→
z =
→
→
b2 ∥ 2
→
b ⋅ b3
∥
→
b3 ∥ 2
−5
= (1, − 1, 1) 0
3
6/9
3
4
7
5
() 3
= (1, − 1, 1) 5 0
4
5
…
4
1
= − 5 + 5 = − 5 and
3
= 5 + 5 = 5,
9/9/2020
Solution →
(
1
7
)
we have ( b) s = (x, y, z) = − 1, − 5 , 5 .
| | { } 1 0 0
7. 7.
Because |A | =
1 1 0
= 1 ≠ 0,
1 1 1
→ → → a 1, a 2, a 3
is a basis of ℝ 3. We use Theorem 7.4.5 to find S 1 and S 2.
→ → 1. Let b 1 = a 1 = (1, 1, 1) and let → → → → a2 ⋅ b1 → b2 = a2 − b 1. → b 1∥ 2
∥
→ → → To find b 2, we need to compute a 2 ⋅ b 1 and
∥
→ → → → b 2 ∥ 2. Because a 2 ⋅ b 1 = (0, 1, 1) · (1, 1, 1) = 2, ∥ b 1 ∥ 2 = 3, and
→ 2 b 2 = (0, 1, 1) − (1, 1, 1) = (0, 1, 1) − 3
(
2 2 2 , , 3 3 3
Let → → → → → → a3 ⋅ b1 → a3 ⋅ b2 → b3 = a3 − b1 − b 2. → → b 1∥ 2 b 2∥ 2
∥
…
7/9
∥
) (
)
2 1 1 = − , , . 3 3 3
9/9/2020
Solution
→ → → → → To find b 3, we need to compute a 3 ⋅ b 1, a 3 ⋅ b 2 and
∥
→ b 2 ∥ 2. Because
(
→ → → → 2 1 1 a 3 ⋅ b 1 = (0, 0, 1) · (1, 1, 1) = 1, a 3 ⋅ b 2 = (0, 0, 1) · − 3 , 3 , 3 1
)
1
= 3 , we obtain
(
→ 1 2 1 1 3 b 3 = (0, 0, 1) − (1, 1, 1) − 1 − , , 3 3 3 3 3
(
)
(
→ → → 2 1 1 1 1 Hence, b 1 = (1, 1, 1), b 2 = − 3 , 3 , 3 , b 3 = 0, − 2 , 2
)
) (
= 0, −
)
1 1 , . 2 2
form an orthogonal basis of ℝ 3.
By computation,
∥
→ b1 ∥ =
→ → → √6 √2 1 → → 1 1 → 3, ∥ b ∥ = , ∥ b = = , b ⋅ b = ( − 1, 1, − 1) · (1, 0, 1) = − 2, b ⋅ b 2 = ( − 1, 1, − 1) · − 2 , 1, 2 √ 2 3 1 3 2 √2
(
∥
→ → and b ⋅ b 3 = ( − 1, 1, − 1) ·
(
2 2
2
, , −3 3 3
(∥
)
2
= 3 . By Theorem 7.4.4,
→ b ⋅ b1
→
→
( b) S = 1
2. By computation, we obtain
…
8/9
→ b 1∥ 2
→ b ⋅ b2
→
,
→ b ⋅ b3
→
,
∥ ∥ → b 2∥ 2
→ b 3∥ 2
)( )( 2
2 1 3 = − , 2, 1 3 3
2
)
2 3 4 . = − , , 3 2 3
)
=1
Solution
→ q1 =
→
b1
∥∥
=
→
1
√3
(1, 1, 1) =
b1 →
b2
∥∥ ∥
→
1 1
,
√6 √6 √6
)
√ 2 √2
forms an orthonormal basis of ℝ 3. By Theorem 7.4.4,
(
)
fo N ot 9/9
1
√2(0, − 2 , 2 ) = (0, − 2 , 2 )
,
∥
→ → → 1 2 → → → → ( b) S = ( b ⋅ q 1, b ⋅ q 2, b ⋅ q 3) = − , , − √2 . 2 3 √ √6
…
1
=
→
b3
2
tri
}
) (
= −
)
is
{
√3 √3 √3
− , . √6 3 3 3
→
b3
1
,
rD
Hence, S 2 =
→ → → q 1, q 2, q 3
(
2 1 1
1
=
b2
→ q3 =
3
,
bu t
→ q2 =
(
1
io n
9/9/2020
9/9/2020
A.8 Eigenvalues and Diagonalizability
A.8 Eigenvalues and Diagonalizability Section 8.1
0
2
0
B = ⎜1
1
0
⎛ )
1
⎝
2
−3 −4
⎛ ⎞ ⎟ ⎠
−1
0
0
0
0
2
0
0
0 −1
0⎟ ⎟ 0⎟
⎞
bu t
1
0
⎜ C = ⎜ ⎜ ⎝
rD is tri
A = (
2
io n
1. For each of the following matrices, find its eigenvalues and determine which eigenvalues are repeated eigenvalues.
0
0
0
0
⎠
2. For each of the following matrices, find its eigenvalues and eigenspaces.
7
3
1
) B = (
3
−1
1
5
) C = (
fo
A = (
5
3
0
0
3
1
2
3
) D = ⎜ 2
1
3⎟
1
2
⎛
⎝
1
⎞
⎠
N
ot
3. Let A be a 3 × 3 matrix. Assume that 2I − A, 3I − A, and 4I − A are not invertible. 1. Find |A| and tr(A). 2. Prove that I + 5A is invertible. 4. Assume that the eigenvalues of a 3 × 3 matrix A are 1, 2, 3. Compute ∣∣A3
…
1/1
2
− 2A
+ 3A + I ∣ ∣
.
9/9/2020
Solution
Solution 1. 1.
By Theorem 8.1.3, the eigenvalues of A are λ 1 = λ 2 = 1, which is a repeated eigenvalue of A. The eigenvalues of B are λ 1 = − 2, λ 2 = 1, and λ 3 = − 4.
Let A =
( ) 3 1
. By computation, we have
p(λ) = |λI − A | =
|
λ − 5 −7 −3
λ−1
|
D is tri b
2. 2.
5 7
ut io n
The eigenvalues of C are λ 1 = − 1, λ 2 = 2, λ 3 = − 2, and λ 4 = 0. The eigenvalue − 1 is a repeated eigenvalue of C.
= (λ − 5)(λ − 1) − 21
= λ 2 − 6λ + 5 − 21 = λ 2 − 6λ − 16 = (λ + 2)(λ − 8).
→
For λ 1 = − 2, (λ 1I − A)→ x = 0 becomes
−3
−7
)( ) ( x
=
−7 −7
N ot
(
−2 − 5
fo r
Solving p(λ) = 0, we obtain λ 1 = − 2 and λ 2 = 8.
−2 − 1
y
−3 −3
)( ) ( ) x y
=
0 0
.
The above system is equivalent to the equation
x + y = 0, where x is a basic variable and y is a free variable. Let y = t. Then x = − t. Hence, Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
1/13
9/9/2020
Solution
() ( ) ( ) x y
{
−t
−1
=t
t
1
→ = tv1,
}
→ t v 1 : t ∈ ℝ .
→
For λ 2 = 8, (λ 2I − A)→ x = 0 becomes −7
−3
8−1
)( ) ( x
=
y
3
−7
−3
7
The above system is equivalent to the equation 3x − 7y = 0,
)( ) ( ) x y
=
0 0
.
D is tri b
(
8−5
ut io n
→ where v 1 = ( − 1, 1). Then E λ = 1
=
7
() x
() 7
→
where v 2 =
3
1
. Then E λ = 2
{
() () 7
7
t
=t 3 1
t 3
→ = tv2,
N ot
y
=
fo r
where x is a basic variable and y is a free variable. Let y = t. Then x = 3 t. Hence,
}
→ tv2 : t ∈ ℝ .
( )
3 −1 Let B = . By computation, Loading [MathJax]/jax/output/HTML-CSS/jax.js 1 5 https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
2/13
9/9/2020
Solution
p(λ) = |λI − B | =
|
λ−3
1
−1
λ−5
|
= (λ − 3)(λ − 5) + 1
Solving p(λ) = 0 we obtain two eigenvalues λ 1 = λ 2 = 4. → For λ 1 = λ 2 = 4, (λ 1I − B)→ x = 0 becomes 1
−1
4−5
)( ) ( x
=
y
1
1
−1 −1
The above system is equivalent to the equation x + y = 0,
)( ) ( ) x y
=
0 0
.
D is tri b
(
4−3
ut io n
= λ 2 − 8λ + 15 + 1 = λ 2 − 8λ + 16 = (λ − 4) 2.
where x is a basic variable and y is a free variable. Let y = t. Then x = − t. Hence,
() ( ) ( ) y
1
Let C =
( ) 3 0 0 3
{
−1
→ = tv1,
=t
t
1
}
N ot
→ where v 1 = ( − 1, 1) T. Then E λ =
=
−t
fo r
x
→ t v 1 : t ∈ ℝ .
. By computation,
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p(λ) = | λI − C | =
|
λ−3
0
0
λ−3
|
= (λ − 3) 2 = 0.
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3/13
9/9/2020
Solution
Hence, λ 1 = λ 2 = 3. → For λ 1 = λ 2 = 3, (λ 1I − C)→ x = 0 becomes x
0 0
y
0
=
0
.
ut io n
( )( ) ( ) 0 0
The corresponding system is 0x + 0y = 0, where x and y are free variables. Let x = s and y = t. Then
() () () () y
=
s t
1
=s
0
+t
→ → T where v 1 = (1, 0) and v 2 = (0, 1) T. Moreover,
( )
{
1 2 3
}
→ → s v 1 + t v 2 : s, t ∈ ℝ .
2 1 3 . By computation, 1 1 2
N ot
Let D =
p(λ) = |λI − D | =
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1
→ → = sv1 + tv2,
fo r
Eλ = Eλ = 1 2
0
D is tri b
x
|
λ−1
−2
−3
−2
λ−1
−3
−1
−1
λ−2
|
= (λ − 1) 2(λ − 2) − 6 − 6 − 3(λ − 1) − 3(λ − 1) − 4(λ − 2) = λ 3 − 4λ 2 − 5λ = λ(λ 2 − 4λ − 5) = λ(λ + 1)(λ − 5).
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4/13
9/9/2020
Solution
Solving p(λ) = λ(λ + 1)(λ − 5) = 0, we get three eigenvalues λ 1 = − 1,
λ 2 = 0, and λ 3 = 5.
→
−2 −2 −3 −1 −1 −3
)( ) ( ) x1
0
x2
=
x3
0 . 0
D is tri b
(
−2 −2 −3
ut io n
For λ 1 = − 1, (λ 1I − E)→ x = 0 becomes
Now we use the row operations to solve the above system.
( |) ( |) ( |) ( |) ( |) ( |) fo r
−2 −2 −3 0 R −1 −1 −3 0 R ( −2) +R 1,3 1 3 −2 −2 −3 0 → −2 −2 −3 0 → −1 −1 −3 0 − 2 − 2 − 3 0 R1 ( − 2 ) + R2
N ot
−1 −1 −3 0 R ( −1) +R −1 −1 −3 0 R (1) +R 2 3 3 1 0 0 3 0 0 0 3 0 → → 0 0 3 0 0 0 0 0 −1 −1 0 0 R ( 1 ) 2 3 0 0 3 0 → 0
0
1 1 0 0
0 0 1 0 . 0 0 R1 ( − 1 ) 0 0 0 0
The system corresponding to the last augmented matrix is Loading [MathJax]/jax/output/HTML-CSS/jax.js https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
5/13
9/9/2020
Solution
{
x1 + x2 = 0 x 3 = 0,
→
For λ 2 = 0, (λ 2I − E)→ x = 0 becomes
−2 −1 −3 −1 −1 −2
)( ) ( ) x1 x2 x3
Now, we use row operations to solve the above system.
0 . 0
|) ( |) ( |) ( |) ( |) fo r
−1 −1 −2 0
R1 ( 2 ) + R2
→
R1 ( 1 ) + R3
R2 ( − 1 ) + R3
→ Loading [MathJax]/jax/output/HTML-CSS/jax.js
=
−1 −2 −3 0 R ( −1) 1 2 3 0 1 −2 −1 −3 0 −2 −1 −3 0 →
N ot
(
0
D is tri b
(
−1 −2 −3
ut io n
where x 1 and x 3 are basic variables and x 2 is a free variable. Let x 2 = t. Then x 1 = − t. Let → v 1 = ( − 1, 1, 0) T. Then
R2 ( − 2 ) + R1
−1 −1 −2 0
1 2 3 0 R (1) 1 2 3 0 2 3 0 3 3 0 → 0 1 1 0 0 1 1 0
0 1 1 0
1 0 1 0
0 1 1 0 . 0 0 0 0
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6/13
9/9/2020
Solution
The system corresponding to the last augmented matrix is
{
x1 + x3 = 0 x 2 + x 3 = 0,
2
{
}
→ t v 2 : t ∈ ℝ .
→
For λ 3 = 5, (λ 3I − E)→ x = 0 becomes
4
−1 −1
−3
)( ) ( ) x1 x2
0
0 . 0
fo r
−2
−2 −3 3
x3
=
N ot
(
4
D is tri b
Eλ =
ut io n
where x 1 and x 2 are basic variables and x 3 is a free variable. Let x 3 = t. Then x 1 = x 2 = − t. Hence, let → v 2 = ( − 1, − 1, 1) T. Then
Now we use row operations to solve the above system.
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7/13
9/9/2020
Solution
|) ( |) |) ( |)
1
1
−2
4
−3 0 R (2) +R 1 3 −3 0 →
1
1
−3 0
0
6
−9 0
− 2 − 3 0 R1 ( − 4 ) + R3 0 − 6
4
R2 ( 1 ) + R3
→
( |)
( |)
( |)
fo r
1 0 −2 0
3 0 . 0 1 −2 0 0 0 0
N ot
→
0
1 1 −3 1 1 −3 0 R ( 1 ) 0 2 6 3 0 6 −9 0 → 0 1 − 0 2 0 0 0 0 0 0 0 0
3
R2 ( − 1 ) + R1
9
ut io n
−2 −3 0 R −1 −1 3 0 R ( −1) 1 , 3 1 −2 4 −3 0 → −2 4 −3 0 → −1 −1 3 0 4 −2 −3 0
D is tri b
( (
4
The system corresponding to the last augmented matrix is
Loading [MathJax]/jax/output/HTML-CSS/jax.js
{
3
x1 − 2 x3 = 0 3
x 2 − 2 x 3 = 0,
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8/13
9/9/2020
Solution
where x 1 and x 3 are basic variables and x 3 is a free variable. Let x 3 = 2t. Then x 1 = 3t and x 2 = 3t. → Hence, Let v 3 = (3, 3, 2) T Then
Eλ =
{
}
ut io n
3
→ t v 3 : t ∈ ℝ
3. 3.
1
D is tri b
1. Because 2I − A, 3I − A and 4I − A are not invertible, |2I − A | = 0, | 3I − A | = 0, and |4I − A | = 0. Hence, 2, 3, and 4 are the eigenvalues of A By Theorem 8.1.4, |A | = 2 × 3 × 4 = 24 and tr(A) = 2 + 3 + 4 = 9.
|
1
2. Because 2, 3, and 4 are the eigenvalues of A, − 5 is not an eigenvalue of A and − 5 I − A Hence,
By Theorem 3.4.1, I + 5A is invertible.
5
I − A) | = ( − 5) 3 | −
1 5
I − A | ≠ 0.
Let ϕ(x) = x 3 − 2x 2 + 3x + 1. Then
N ot
4. 4.
1
≠ 0.
fo r
|I + 5A | = | − 5( −
|
ϕ(A) = A 3 − 2A 2 + 3A + 1.
Because 1, 2, 3 are the eigenvalues of A, it follows from Theorem 8.1.5 that ϕ(1), ϕ(2), and ϕ(3) are all eigenvalues of ϕ(A). By computation, ϕ(1) = 1 − 2 + 3 + 1 = 3, ϕ(2) = 2 3 − 2(2) 2 + 3(2) + 1 = 7, and ϕ(3) = 3 3 − 2(3) 2 + 3(3) + 1 = 19.
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9/13
9/9/2020
Solution
|
By Theorem 8.1.4, ϕ(A) | = | A 3 − 2A 2 + 3A + I | = 3 × 7 × 19 = 399.
Section 8.2
(
2
4
0
0
3
1
7
0
0
5
8
0
0
0
−2
)
2. Find x ∈ ℝ such that A is diagonalizable, where
(
→
2
0
−4
1
1
0
4
0
−1
0
−4
3
)
1 . 3
1
)
x . 2
fo r
A=
(
B=
−1
D is tri b
A=
−1
ut io n
1. For each of the following matrices, determine whether it is diagonalizable.
N ot
3. Assume that υ = (1, 1, 0) T is an eigenvector of the matrix
A=
(
1
−1
3
x
y
0
1
)
2 . 1 →
i. Find x, y ∈ ℝ R and the eigenvalue corresponding to υ . ii. Is A diagonalizable?
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10/13
9/9/2020
Solution
4. Let
(
A=
1
−1
−4
1
)
.
ut io n
1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1. 5. Let
( ) 2
4
2
0
4
2
2 . 3
D is tri b
A=
3
fo r
1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1.
N ot
3. Compute A 2 by using the result (2). 6. Let
A=
(
0
0
−2
1
2
1
1
0
3
)
.
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1. Find the eigenvalues and eigenspaces of A. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
11/13
9/9/2020
Solution
2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1. 3. Compute A 2 by using the result (2). 7. Let
(
)
−1
1
3
−3
0
0
2 . 1
ut io n
A=
1
3. Compute A 3 by using the result (2). 8. Let
(
1
−1
0
)
.
fo r
A=
0
D is tri b
1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1.
1. Find the eigenvalues and eigenspaces of A.
N ot
2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1. 3. Compute A 2 by using the result (2).
9. Two n × n matrices A and B are said to be similar (or A is similar to B) if there exists an invertible n × n matrix P such that B = P − 1AP. We write A~B. Let
A= Loading [MathJax]/jax/output/HTML-CSS/jax.js
(
1
−1
2
1
)
,
B=
(
1
−2
1
1
)
, and P =
(
0
1
−1
0
)
.
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12/13
Solution
Show that A is similar to B. 10. Assume that A~B. Show that the following assertions hold. i. |λI − A | = | λI − B | . ii. A and B have the same eigenvalues. iii. tr(A) = tr(B). iv. |A | = | B | . v. r(A) = r(B).
ut io n
9/9/2020
D is tri b
11. Assume that A, B, C are n × n matrices. Show that the following assertions hold i. A~A. ii. If A~B, then B~A. iii. If A~B and B~C, then A~C.
12. Assume that A and B are n × n matrices and A~B. Prove that the following assertions hold. a. A k~B k for every positive integer k.
N ot
fo r
b. Let ϕ(λ) = a mλ m + a m − 1λ m − 1 + ⋯ + a 1λ + a 0. Then ϕ(A)~ϕ(B).
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13/13
9/9/2020
Solution
Solution 1. 1.
By Theorem 8.1.3, the eigenvalues of A are λ 1 = λ 2 = 1, which is a repeated eigenvalue of A. The eigenvalues of B are λ 1 = − 2, λ 2 = 1, and λ 3 = − 4.
Let A =
( ) 3 1
. By computation, we have
p(λ) = |λI − A | =
|
λ − 5 −7 −3
λ−1
|
D is tri b
2. 2.
5 7
ut io n
The eigenvalues of C are λ 1 = − 1, λ 2 = 2, λ 3 = − 2, and λ 4 = 0. The eigenvalue − 1 is a repeated eigenvalue of C.
= (λ − 5)(λ − 1) − 21
= λ 2 − 6λ + 5 − 21 = λ 2 − 6λ − 16 = (λ + 2)(λ − 8).
→
For λ 1 = − 2, (λ 1I − A)→ x = 0 becomes
−3
−7
)( ) ( x
=
−7 −7
N ot
(
−2 − 5
fo r
Solving p(λ) = 0, we obtain λ 1 = − 2 and λ 2 = 8.
−2 − 1
y
−3 −3
)( ) ( ) x y
=
0 0
.
The above system is equivalent to the equation
x + y = 0, where x is a basic variable and y is a free variable. Let y = t. Then x = − t. Hence, Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
1/13
9/9/2020
Solution
() ( ) ( ) x y
{
−t
−1
=t
t
1
→ = tv1,
}
→ t v 1 : t ∈ ℝ .
→
For λ 2 = 8, (λ 2I − A)→ x = 0 becomes −7
−3
8−1
)( ) ( x
=
y
3
−7
−3
7
The above system is equivalent to the equation 3x − 7y = 0,
)( ) ( ) x y
=
0 0
.
D is tri b
(
8−5
ut io n
→ where v 1 = ( − 1, 1). Then E λ = 1
=
7
() x
() 7
→
where v 2 =
Let B = Processing math: 100%
3
. Then E λ =
1
2
{
() () 7
7
t
=t 3 1
t 3
→ = tv2,
N ot
y
=
fo r
where x is a basic variable and y is a free variable. Let y = t. Then x = 3 t. Hence,
}
→ tv2 : t ∈ ℝ .
( ) 3 −1 1
5
. By computation,
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2/13
9/9/2020
Solution
p(λ) = |λI − B | =
|
λ−3
1
−1
λ−5
|
= (λ − 3)(λ − 5) + 1
Solving p(λ) = 0 we obtain two eigenvalues λ 1 = λ 2 = 4. → For λ 1 = λ 2 = 4, (λ 1I − B)→ x = 0 becomes 1
−1
4−5
)( ) ( x
=
y
1
1
−1 −1
The above system is equivalent to the equation x + y = 0,
)( ) ( ) x y
=
0 0
.
D is tri b
(
4−3
ut io n
= λ 2 − 8λ + 15 + 1 = λ 2 − 8λ + 16 = (λ − 4) 2.
where x is a basic variable and y is a free variable. Let y = t. Then x = − t. Hence,
() ( ) ( ) y
1
Let C =
Processing math: 100%
( ) 3 0 0 3
{
−1
→ = tv1,
=t
t
1
}
N ot
→ where v 1 = ( − 1, 1) T. Then E λ =
=
−t
fo r
x
→ t v 1 : t ∈ ℝ .
. By computation,
p(λ) = | λI − C | =
|
λ−3
0
0
λ−3
|
= (λ − 3) 2 = 0.
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3/13
9/9/2020
Solution
Hence, λ 1 = λ 2 = 3. → For λ 1 = λ 2 = 3, (λ 1I − C)→ x = 0 becomes x
0 0
y
0
=
0
.
ut io n
( )( ) ( ) 0 0
The corresponding system is 0x + 0y = 0, where x and y are free variables. Let x = s and y = t. Then
() () () () y
=
s t
1
=s
0
+t
→ → T where v 1 = (1, 0) and v 2 = (0, 1) T. Moreover,
( )
{
1 2 3
2 1 3 . By computation, 1 1 2
p(λ) = |λI − D | =
Processing math: 100%
}
→ → s v 1 + t v 2 : s, t ∈ ℝ .
N ot
Let D =
1
→ → = sv1 + tv2,
fo r
Eλ = Eλ = 1 2
0
D is tri b
x
|
λ−1
−2
−3
−2
λ−1
−3
−1
−1
λ−2
|
= (λ − 1) 2(λ − 2) − 6 − 6 − 3(λ − 1) − 3(λ − 1) − 4(λ − 2) = λ 3 − 4λ 2 − 5λ = λ(λ 2 − 4λ − 5) = λ(λ + 1)(λ − 5).
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4/13
9/9/2020
Solution
Solving p(λ) = λ(λ + 1)(λ − 5) = 0, we get three eigenvalues λ 1 = − 1,
λ 2 = 0, and λ 3 = 5.
→
−2 −2 −3 −1 −1 −3
)( ) ( ) x1
0
x2
=
x3
0 . 0
D is tri b
(
−2 −2 −3
ut io n
For λ 1 = − 1, (λ 1I − E)→ x = 0 becomes
Now we use the row operations to solve the above system.
( |) ( |) ( |) ( |) ( |) ( |) fo r
−2 −2 −3 0 R −1 −1 −3 0 R ( −2) +R 1,3 1 3 −2 −2 −3 0 → −2 −2 −3 0 → −1 −1 −3 0 − 2 − 2 − 3 0 R1 ( − 2 ) + R2
N ot
−1 −1 −3 0 R ( −1) +R −1 −1 −3 0 R (1) +R 2 3 3 1 0 0 3 0 0 0 3 0 → → 0 0 3 0 0 0 0 0 −1 −1 0 0 R ( 1 ) 2 3 0 0 3 0 → 0
0
1 1 0 0
0 0 1 0 . 0 0 R1 ( − 1 ) 0 0 0 0
The system corresponding to the last augmented matrix is Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
5/13
9/9/2020
Solution
{
x1 + x2 = 0 x 3 = 0,
→
For λ 2 = 0, (λ 2I − E)→ x = 0 becomes
−2 −1 −3 −1 −1 −2
)( ) ( ) x1 x2 x3
Now, we use row operations to solve the above system.
0 . 0
|) ( |) ( |) ( |) ( |) fo r
−1 −1 −2 0
R1 ( 2 ) + R2
→
R1 ( 1 ) + R3
R2 ( − 1 ) + R3
→ Processing math: 100%
=
−1 −2 −3 0 R ( −1) 1 2 3 0 1 −2 −1 −3 0 −2 −1 −3 0 →
N ot
(
0
D is tri b
(
−1 −2 −3
ut io n
where x 1 and x 3 are basic variables and x 2 is a free variable. Let x 2 = t. Then x 1 = − t. Let → v 1 = ( − 1, 1, 0) T. Then
R2 ( − 2 ) + R1
−1 −1 −2 0
1 2 3 0 R (1) 1 2 3 0 2 3 0 3 3 0 → 0 1 1 0 0 1 1 0
0 1 1 0
1 0 1 0
0 1 1 0 . 0 0 0 0
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6/13
9/9/2020
Solution
The system corresponding to the last augmented matrix is
{
x1 + x3 = 0 x 2 + x 3 = 0,
2
{
}
→ t v 2 : t ∈ ℝ .
→
For λ 3 = 5, (λ 3I − E)→ x = 0 becomes
4
−1 −1
−3
)( ) ( ) x1 x2
0
0 . 0
fo r
−2
−2 −3 3
x3
=
N ot
(
4
D is tri b
Eλ =
ut io n
where x 1 and x 2 are basic variables and x 3 is a free variable. Let x 3 = t. Then x 1 = x 2 = − t. Hence, let → v 2 = ( − 1, − 1, 1) T. Then
Now we use row operations to solve the above system.
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7/13
9/9/2020
Solution
|) ( |) |) ( |)
1
1
−2
4
−3 0 R (2) +R 1 3 −3 0 →
1
1
−3 0
0
6
−9 0
− 2 − 3 0 R1 ( − 4 ) + R3 0 − 6
4
R2 ( 1 ) + R3
→
( |)
( |)
( |)
fo r
1 0 −2 0
3 0 . 0 1 −2 0 0 0 0
N ot
→
0
1 1 −3 1 1 −3 0 R ( 1 ) 0 2 6 3 0 6 −9 0 → 0 1 − 0 2 0 0 0 0 0 0 0 0
3
R2 ( − 1 ) + R1
9
ut io n
−2 −3 0 R −1 −1 3 0 R ( −1) 1 , 3 1 −2 4 −3 0 → −2 4 −3 0 → −1 −1 3 0 4 −2 −3 0
D is tri b
( (
4
The system corresponding to the last augmented matrix is
Processing math: 100%
{
3
x1 − 2 x3 = 0 3
x 2 − 2 x 3 = 0,
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8/13
9/9/2020
Solution
where x 1 and x 3 are basic variables and x 3 is a free variable. Let x 3 = 2t. Then x 1 = 3t and x 2 = 3t. → Hence, Let v 3 = (3, 3, 2) T Then
Eλ =
{
}
ut io n
3
→ t v 3 : t ∈ ℝ
3. 3.
1
D is tri b
1. Because 2I − A, 3I − A and 4I − A are not invertible, |2I − A | = 0, | 3I − A | = 0, and |4I − A | = 0. Hence, 2, 3, and 4 are the eigenvalues of A By Theorem 8.1.4, |A | = 2 × 3 × 4 = 24 and tr(A) = 2 + 3 + 4 = 9.
|
1
2. Because 2, 3, and 4 are the eigenvalues of A, − 5 is not an eigenvalue of A and − 5 I − A Hence,
By Theorem 3.4.1, I + 5A is invertible.
5
I − A) | = ( − 5) 3 | −
1 5
I − A | ≠ 0.
Let ϕ(x) = x 3 − 2x 2 + 3x + 1. Then
N ot
4. 4.
1
≠ 0.
fo r
|I + 5A | = | − 5( −
|
ϕ(A) = A 3 − 2A 2 + 3A + 1.
Because 1, 2, 3 are the eigenvalues of A, it follows from Theorem 8.1.5 that ϕ(1), ϕ(2), and ϕ(3) are all eigenvalues of ϕ(A). By computation, ϕ(1) = 1 − 2 + 3 + 1 = 3, ϕ(2) = 2 3 − 2(2) 2 + 3(2) + 1 = 7, and ϕ(3) = 3 3 − 2(3) 2 + 3(3) + 1 = 19.
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9/13
9/9/2020
Solution
|
By Theorem 8.1.4, ϕ(A) | = | A 3 − 2A 2 + 3A + I | = 3 × 7 × 19 = 399.
Section 8.2
(
2
4
0
0
3
1
7
0
0
5
8
0
0
0
−2
)
2. Find x ∈ ℝ such that A is diagonalizable, where
(
→
2
0
−4
1
1
0
4
0
−1
0
−4
3
)
1 . 3
1
)
x . 2
fo r
A=
(
B=
−1
D is tri b
A=
−1
ut io n
1. For each of the following matrices, determine whether it is diagonalizable.
N ot
3. Assume that υ = (1, 1, 0) T is an eigenvector of the matrix
A=
(
1
−1
3
x
y
0
1
)
2 . 1 →
i. Find x, y ∈ ℝ R and the eigenvalue corresponding to υ . ii. Is A diagonalizable?
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10/13
9/9/2020
Solution
4. Let
(
A=
1
−1
−4
1
)
.
ut io n
1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1. 5. Let
( ) 2
4
2
0
4
2
2 . 3
D is tri b
A=
3
fo r
1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1.
N ot
3. Compute A 2 by using the result (2). 6. Let
A=
(
0
0
−2
1
2
1
1
0
3
)
.
Processing math: 100%
1. Find the eigenvalues and eigenspaces of A. https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
11/13
9/9/2020
Solution
2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1. 3. Compute A 2 by using the result (2). 7. Let
(
)
−1
1
3
−3
0
0
2 . 1
ut io n
A=
1
3. Compute A 3 by using the result (2). 8. Let
(
1
−1
0
)
.
fo r
A=
0
D is tri b
1. Find the eigenvalues and eigenspaces of A. 2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1.
1. Find the eigenvalues and eigenspaces of A.
N ot
2. Find an inverse matrix P and a diagonal matrix D such that A = PDP − 1. 3. Compute A 2 by using the result (2).
9. Two n × n matrices A and B are said to be similar (or A is similar to B) if there exists an invertible n × n matrix P such that B = P − 1AP. We write A~B. Let
A= Processing math: 100%
(
1
−1
2
1
)
,
B=
(
1
−2
1
1
)
, and P =
(
0
1
−1
0
)
.
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12/13
Solution
Show that A is similar to B. 10. Assume that A~B. Show that the following assertions hold. i. |λI − A | = | λI − B | . ii. A and B have the same eigenvalues. iii. tr(A) = tr(B). iv. |A | = | B | . v. r(A) = r(B).
ut io n
9/9/2020
D is tri b
11. Assume that A, B, C are n × n matrices. Show that the following assertions hold i. A~A. ii. If A~B, then B~A. iii. If A~B and B~C, then A~C.
12. Assume that A and B are n × n matrices and A~B. Prove that the following assertions hold. a. A k~B k for every positive integer k.
N ot
fo r
b. Let ϕ(λ) = a mλ m + a m − 1λ m − 1 + ⋯ + a 1λ + a 0. Then ϕ(A)~ϕ(B).
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13/13
9/9/2020
Solution
Solution 1. 1. Because A is a 4 × 4 upper triangular matrix, by Theorem 8.1.3, all the eigenvalues of A are the entries on the main diagonal. Hence, λ 1 = − 1, λ 2 = 3, λ 3 = 5, and λ 4 = − 2 are the eigenvalues of A
=
−4
−3
0
λ+1
−1
0
4
λ−3
Then λ 1 = − 1, λ 2 = λ 3 = 1. For λ 2 = λ 3 = 1, the system λ 2I − A = 0 becomes
|
= (λ + 1)(λ − 1) 2 = 0.
D is tri b
| || λI − B
λ+1
ut io n
and A has four distinct eigenvalues. It follows from Corollary 8.2.1 (1) that A is diagonalizable.
( )( ) ( ) x1
0 2 −1
x2
0 4 −2
x3
0
fo r
2 4 −3
0 . 0
N ot
We solve the above system.
=
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1/27
9/9/2020
Solution
→
(A | 0) :
=
( |) ( |) 2 4 −3 0 R +R −2 2 0 2 −1 0 →
2 0 −1 0 0 2 −1 0
0 4 − 2 0 R2 ( − 2 ) + R1 0 0
( |)
0
0
1
R1 ( 2 )
→ 1
1 0 . 0 1 −2 0 0 0 0
D is tri b
R1 ( 2 )
1 0 −2 0
ut io n
1
The system corresponding to the last augmented matrix has one free variable and thus, dim(E λ ) = 1 < 2 (algebraic multiplicity). By Corollary 8.2.1 (ii), A is not diagonalizable. 2
| || λI − A
=
0
−1
4
λ−1
−x
−1
0
λ−2
|
= (λ − 1) 2(λ − 3) = 0.
fo r
2. 2.
λ−2
Then λ 1 = 1 (with algebraic multiplicity 2), λ 2 = 3.
N ot
We find dim(Eλ 1). For λ 1 = 1, the system λ 1I − A = 0 becomes
( )( ) ( ) −1 0 −1
x1
0 −x
x2
−1 0 −1
x3
4
0
=
0 . 0
We solve the above system.
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2/27
9/9/2020
Solution
→
(A | 0) =
( |) ( −1 0 −1 0 R (4) +R 1 2 4 0 −x 0 →
− 1 0 − 1 0 R1 ( − 1 ) + R3
−1 0 0 0
|)
−1
0
0 −x − 4 0 . 0 0 0
3. 3.
ut io n
If x = − 4, then the system corresponding to the last augmented matrix has two free variables and dim(Eλ 1) = 2 (algebraic multiplicity). By Corollary 8.2.1 (i), A is diagonalizable. →
i. Let λ ∈ ℝ be the eigenvalue of A corresponding to → v. Then (λI − A)→ v = 0, that is, 1
−1
−3
λ−x
−2
−y
0
λ−1
This implies that
)( ) ( ) 1 1
0
=
0 . 0
D is tri b
(
λ−1
0
( ) () λ
0 0
fo r
λ−x−3
=
0
N ot
−y
and λ = 0, λ − x − 3 = 0, and − y = 0. Hence, x = − 3, y = 0, and λ = 0. ii. When x = − 3, y = 0 and λ = 0, A becomes
( ) 1 −1 1
A=
3 −3 2 . 0 0 1
Processing math: 100%
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3/27
9/9/2020
Solution
|λI − A | =
|
λ−1
1
−1
−3
λ+3
−2
0
0
λ−1
|
= λ(λ − 1)(λ + 2) = 0.
ut io n
Then λ 1 = − 2, λ 2 = 0, and λ 3 = 1. By Corollary 8.2.1 (1), A is diagonalizable. 4. 4.
p(λ) = | λI − A | =
|
λ−1
1
4
λ−1
|
= (λ + 1)(λ − 3).
D is tri b
1.
Solving p(λ) = (λ + 1)(λ − 3) = 0 implies λ 1 = − 1 and λ 2 = 3. →
→
For λ 1 = − 1, (λ 1I − A)X = 0 becomes 1
4
−1 − 1
)( ) ( x y
=
−2
1
4
−2
fo r
(
−1 − 1
)( ) ( ) x y
=
0 0
.
N ot
The above system is equivalent to the equation
− 2x + y = 0,
where x is a basic variable and y is a free variable. Let y = 2t. Then x = t. Hence,
() ( ) () x y
=
t
2t
=t
1 2
→ = tv1,
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4/27
9/9/2020
Solution
→ where v 1 = (1, 2) T. Then E λ = 1 →
{
}
→ tv1 : t ∈ ℝ .
→
For λ 2 = 3, (λ 2I − A)X = 0 becomes 1
4
3−1
)( ) ( )( ) ( ) x
=
y
2 1
x
4 2
y
=
The above system is equivalent to
0
.
D is tri b
2x + y = 0,
0
ut io n
(
3−1
where x is a basic variable and y is a free variable. Let y = − 2t. Then x = t and
() ( ) ( ) x y
2. Let
1
−2
= tv 2,
}
→ tv2 : t ∈ ℝ .
fo r
{
=t
− 2t
N ot
→ where v 2 = (1, − 2) T. Then E λ = 2
=
t
( )
→→ 1 1 D = diag(λ 1, λ 2) = diag( − 1, 3) and P = ( v 1 v 2 ) = . 2 −2 Then by Theorem 8.2.2, A = P − 1DP. 5. 5.
Processing math: 100%
1. Because
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5/27
9/9/2020
Solution
p(λ) = |λI − A | =
|
λ − 3 −2 −2 −4
−4
λ
−2
−2 λ − 3
|
= λ 3 − 6λ 2 + 15λ − 8
= (λ + 1) 2(λ − 8) = 0.
ut io n
Then λ 1 = − 1 (with algebraic multiplicity 2), λ 2 = 8. We find dim(Eλ 1). For λ 1 = 1, the system λ 1I − A = 0 becomes
−2 −1 −2 −4 −2 −4
)( ) ( )
|
→
(
x2
=
x3
0 . 0
|) (
|)
−4 −2 −4 0 R ( − 1 ) +R −4 −2 −4 0 1 2 2 −2 −1 −2 0 0 0 0 0 → − 4 − 2 − 4 0 R1 ( − 1 ) + R3
N ot
(A 0) =
0
fo r
We solve the above system.
x1
D is tri b
(
−4 −2 −4
0
0
0
0
( |) 1
1
R 1( − 4 )→
1 2 1 0 0 . 0 0 0 0 0 0 0
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6/27
9/9/2020
Solution
The system with the last augmented matrix is equivalent to the equation 1
x+
y + z = 0,
2
ut io n
where x is a basic variable and y, z are free variables. Let y = 2s and z = t. Then x = − s − t. Hence,
() ( ) ( ) ( ) x
−s − t
y
2s
=
2
=s
t
−1
+t
0
→ → = tv1 + sv2,
0 1
D is tri b
z
−1
→ → T where v 1 = ( − 1, 2, 0) and v 2 = ( − 1, 0, 1) T. Then
Eλ =
{
}
fo r
1
→ → s v 1 + t v 2 : s, t ∈ ℝ .
N ot
For λ 2 = 8, the system λ 2I − A = 0 becomes
(
5
−2
−2 −4 8
−4 −2
−2 5
)( ) ( ) x1 x2 x3
0
=
0 . 0
We solve the above system. Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml/f…
7/27
9/9/2020
Solution
(A | 0)
|) ( |) ( |) ( |) ( |) =
(
5
−2
|)
−2 −4 0
− 2 0 R (2) + R → 2 1 −4 −2 5 0
R1 ( 2 ) + R2
→
R1 ( 4 ) + R3
8
(
1
14 − 8 0
−2
8
−2 0
−4 −2
5
0
− 8 0 R ( 1 ) 1 14 − 8 0 2 18 0 36 − 18 0 0 2 −1 0 → 1 14
1
0 54 − 27 0 R 3 ( ) 0 27 1 14 − 8 0 0
2
0
0
1 0 −1 0
− 1 0 R ( − 7) + R → 3 1 0 0
fo r
0 2 −1 0 . 0 0 0 0
−1 0
D is tri b
R 2( − 1) + R 3→
2
ut io n
→
The system with the last augmented matrix is equivalent to the system
N ot
{
x−z=0 2y − z = 0,
where x and y are basic variables and z is a free variable. → Let z = 2t. Then x = 2t and y = t. Let v 3 = (2, 1, 2) T. Then
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8/27
9/9/2020
Solution
Eλ = 2
{
}
→ tv3 : t ∈ ℝ .
Then by Theorem 8.2.2, A = PDP − 1.
| ( | ) ( | ) ( | ) ( | ) −1 −1 2 1 0 0
(P I) =
2
0
0
1
0
1
1 . 2
−2 5 2 1 0 R → 2, 3 1 2 0 0 1
N ot
0
0
1 0 1 0 R (2) + R → 1 3 2 0 0 1
−1 −1 2 1 0 0 0
2
fo r
3.
( ) −1 −1 2
D is tri b
→→→ P = (v1 v2 v3) =
ut io n
2. Let D = diag(λ 1, λ 1, λ 2) = diag( − 1, − 1, 8) and
−1 −1 2 1 0 0 0 0
1
2 0 0 1 R (2) + R → 2 3 −2 5 2 1 0
−1 −1 2 1 0 0
Processing math: 100%
0
1
0
0
1
2 0 0 0 R ( )→ 3 9 9 2 1 2
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9/27
9/9/2020
Solution
( | ) −1 −1 2 0
1
0
0
2 1
1 0 0 0 0 1 2
1
2
9
9
9
R3 ( − 2 ) + R2
→
R3 ( − 2 ) + R1
1
0
0
4
2
9
1 9
−1 0 0
4
4
−9 2
9
|
R 2(1) + R 1→
2
9
1 0 − −9 9 0 1 2 1
0
9 9
1 9 5 9 2
1
1 0 0
R 1( − 1)→
N ot
0
5
0 − −9 9 1 2 1
D is tri b
0
4
−9 −9
fo r
9
−1 −1 0
2
ut io n
( | ) ( | ) (| ) 5
9
9
−9 4
4 9
2
0 1 0 − −9 9 0 0 1 2 1 9
9
1
−9 5 9 2 9
= (I P − 1). Hence, Processing math: 100%
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10/27
Solution
( ) 1
−9
P −1 =
4
4 9
2
−9 −9
1
−9 5 9
2
1
2
9
9
9
N ot
fo r
D is tri b
By Theorem 8.2.6,
.
ut io n
9/9/2020
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11/27
Solution
( )( ) −1 −1 2
A 2 = PD 2P − 1 =
0
1
0
0
1
2
0
( ) 4
9 0 2 4 2 −1 0 −9 −9 0 8 2 1
−1
2
1
−9
0
9
( )( ) ( ) ( )( ) ( ) ( ) −1 −1 2 2
0
1
0
1
2
4
0 ][ − 4 − 2 9 0 0 64 2 1
2
0
64
0
1
128
−1
4
−4 −2 2
1
−1 5
]
9 2 9
2
−1 5 2
fo r
= 9
−1
1
0 1
− 1 − 1 128
1
0
5
D is tri b
= [
1 0
9
1
−9
ut io n
9/9/2020
261 126 252
1
N ot
= 9 126 72 126 252 126 261 29 14 28
=
14
8
14 . 28 14 29
6. 6. Processing math: 100%
1.
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12/27
9/9/2020
Solution
|
λ
0
|
2
p(λ) = |λI − A | = − 1 λ − 2 − 1 = λ(λ − 2)(λ − 3) + 2(λ − 2) −1 0 λ−3 = (λ − 2)(λ 2 − 3λ + 2) = (λ − 2)(λ − 2)(λ − 1)
ut io n
= (λ − 1)(λ − 2) 2 = 0. This implies λ 1 = 1 and λ 2 = λ 3 = 2. For λ 1 = 1, the system λ 1I − A = 0 becomes
−1 −1 −1 −1
0
−1
0
−2
x2
0
=
x3
0 . 0
|) ( |) ( |) 2
0 R (1) +R 1 0 2 0 1 2 −1 −1 −1 0 0 −1 1 0 →
fo r
1
0
)( ) ( ) x1
2
− 2 0 R1 ( 1 ) + R3 0
N ot
(
0
D is tri b
(
1
1 0
R 2( − 1)→
2
0
0 0
0
0 1 −1 0 . 0 0 0 0
The system corresponding to the last augmented matrix is Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
13/27
9/9/2020
Solution
{
x 1 + 2x 3 = 0 x 2 − x 3 = 0,
1
{
→ tv1 : t ∈ ℝ
}
For λ 2 = λ 3 = 2, the system λ 2I − A = 0 becomes
→ = span{ v 1 }.
D is tri b
Eλ =
ut io n
where x 1 and x 2 are basic variables and x 3 is a free variable. Let x 3 = t. Then x 1 = − 2t and → x 2 = t. Let v 1 = ( − 2, 1, 1) T. We have
( )( ) ( ) 2
0
x1
2
x2
−1 0 −1
x3
=
fo r
−1 0 −1
0
0 . 0
( |) ( |) ( |) 0
2
0
N ot
2
1
− 1 0 − 1 0 R 1( )→ 2 −1 0 −1 0
R1 ( 1 ) + R2
→
R1 ( 1 ) + R3
1
0
1
0
−1 0 −1 0 −1 0 −1 0
1 0 1 0
0 0 0 0 . 0 0 0 0
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14/27
9/9/2020
Solution
The system corresponding to the last augmented matrix is x 1 + x 3 = 0, where x 1 is a basic variable, and x 2 and x 3 are free variables. Let x 2 = s and x 3 = t. Then
ut io n
x 1 = − t. Then (x 1, x 2, x 3) = ( − t, s, t) = ( − t, 0, t) + (0, s, 0) = t( − 1, 0, 1) + s(0, 1, 0).
Eλ = 2
D is tri b
→ → Hence, we get two basic vectors v 2 = ( − 1, 0, 1) T and v 3 = (0, 1, 0) T. Hence,
{
→ → s v 2 + t v 3 : s, t ∈ ℝ
( )( ) λ1
0
0
0
λ2
0
0
λ3
N ot
D=
0
and
Processing math: 100%
fo r
2. Let
} { }
→ → = span v 2 , v 3 .
→→→ P = (v1 v2 v3) =
1 0 0
=
0 2 0 0 0 2
( ) −2 −1 0 1
0
1
1
1 . 0
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9/9/2020
Solution
Then P diagonalizes A and by Theorem 8.2.2, A = PDP − 1. 3.
( | ( ( | (
) ( | ) | ) ) ( | ) | )
−2 −1 0 1 0 0
=
1
0
1
1
1
R1 ( 2 ) + R1
R1 ( − 1 ) + R3
0
0
0
1
1
0
1
2
1
2
1 0 1 0
−2 −1 0 1 0 0 1
1
0 0 0 1
0
0 R 2(1) + R 3→ −1 0 −1 1
1 0 1 0 R ( −2) +R 1 0 0 −1 0 −1 3 2 0 −1 2 1 2 0 0 −1 0 −1 0 −2 →
D is tri b
0
0
0 −1
→
1
1 0 1 0 R 1 , 2→ 0 0 0 1
0
ut io n
(P | I)
1
1 1 1 1 R3 ( − 1 ) + R1 0
0
1
1
1
1
1 0 0 −1 0 −1 0 1 0
0
2
1
1
= (I | P − 1)
1
N ot
0 0 1
Hence,
1
fo r
R 2( − 1)→
1
0
2
1
1
1
( ) −1 0 −1
P −1 =
.
By Theorem 8.2.6, Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
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9/9/2020
Solution
( )( ) ( ) ( )( )( ) ( )( ) ( ) −2 −1 0 1
0
1
1
1
0
0 0 2
−2 −1 0 1
0
1
1
1
0
=
0
4
1
0
2
1
4
0
1
1
7. 7.
|
0
2
1
1
1
−2 0 −6 3
4
3
1
3
0
7
|
|
=
λ−1
1
−1
−3
λ+3
−2
0
0
λ−1
=λ−1
fo r
p(λ) = |λI − A | =
1
−1 0 −1
1
i.
1
−1 0 −1
0 4 0 ] 0 0 4
−2 −4 0
1
.
D is tri b
= [
1 0 0
1
ut io n
A 2 = PD 2P − 1 =
1 0 0 2 −1 0 −1 0 2 0 1 0 2
λ−1
1
−3
λ+3
|
N ot
= (λ − 1)[(λ − 1)(λ + 3) − ( − 3)] = (λ − 1)(λ 2 + 2λ) = (λ + 2)λ(λ − 1). This implies λ 1 = − 2, λ 2 = 0, and λ 3 = 1. For λ 1 = − 2, the system λ 1I − A = 0 becomes
Processing math: 100%
( )( ) ( ) −3 1 −1
x1
−3 1 −2
x2
0
0 −3
x3
0
=
0 . 0
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9/9/2020
Solution
( |) ( |) ( |) ( |) −3 1 −1 0
− 3 1 − 2 0 R ( − 1) + R → 1 2 0 0 −3 0
0
0 −1 0
0
0 −3 0
−3 1 −1 0
R 2( − 3) + R 3→
0
−3 1
0 − 1 0 R ( − 1) + R → 3 1 0 0 0
0
{
0
0 −1 0 . 0 0 0
D is tri b
The system corresponding to the last augmented matrix is
0
0 0
ut io n
−3 1 −1 0
− 3x 1 + x 2 = 0 − x 3 = 0,
{
→ tv1 : t ∈ ℝ
N ot
Eλ =
fo r
where x 1 and x 2 are basic variables and x 2 is a free variable. Let x 2 = 3t. Then x 1 = t and → x 3 = 0. Let v 1 = (1, 3, 0) T. We have
1
} {}
→ = span v 1 .
For λ 2 = 0, the system λ 2I − A = 0 becomes
Processing math: 100%
( )( ) ( ) −1 1 −1
x1
−3 3 −2
x2
0
0 −1
x3
0
=
0 . 0
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18/27
9/9/2020
Solution
( |) ( |) ( |) −1 1 −1 0
−1 1 −1 0
− 3 3 − 2 0 R ( − 3) + R → 1 2 0 0 −1 0
0
1 0
0
0 −1 0
−1 1 0 0
→
0
R2 ( 1 ) + R1
0
0 1 0 . 0 0 0
{
D is tri b
The system corresponding to the last augmented matrix is
ut io n
R2 ( 1 ) + R2
0
− x 1 + x 2 = 0, x 3 = 0,
fo r
where x 1 and x 3 are basic variables and x 2 is a free variable. Let x 2 = t. Then x 1 = t. Then → (x 1, x 2, x 3) = (t, t, 0) = t(1, 1, 0). Let v 2 = (1, 1, 0) T. We have
2
{
} {}
→ = span v 2 .
N ot
Eλ =
→ tv2 : t ∈ ℝ
For λ 3 = 1, the system λ 3I − A = 0 becomes
( )( ) ( ) 1 −1
x1
−3 4 −2
x2
0
Processing math: 100%
0
0
0
x3
0
=
0 . 0
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9/9/2020
Solution
( |) ( |) ( |) 1 −1 0
−3 4 −2 0
−3 4 −2 0 R → 1, 2 0 0 0 0 −3 0 0 0
2
0
1 − 1 0 R ( − 4) + R → 2 1 0 0 0
0
0
1 −1 0 . 0 0 0
{
D is tri b
The system corresponding to the last augmented matrix is
ut io n
0
− 3x 1 + 2x 3 = 0, x 2 − x 3 = 0,
3
ii. Let
D= Processing math: 100%
{
→ tv3 : t ∈ ℝ
N ot
Eλ =
fo r
where x 1 and x 2 are basic variables and x 23 is a free variable. Let x 3 = 3t. Then x 1 = 2t and → x 2 = 3t. Then (x 1, x 2, x 3) = (2t, 3t, 3) = t(2, 3, 3). Let v 2 = (2, 3, 3) T. We have
(
λ1
0
0
0
λ2
0
0
0
λ3
} {}
→ = span v 3 .
)( ) −2 0 0
=
0
0 0
0
0 1
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9/9/2020
Solution
and
( ) 1 1 2
3 1 3 . 0 0 3
Then P diagonalizes A and by Theorem 8.2.2, A = PDP − 1. iii.
( | ) 1 1 2 1 0 0
=
3 1 3 0 1 0 R ( − 3) + R → 1 2 0 0 3 0 0 1
( | ) 1
1
1
2
0 −2 −3 0
0
0
2
1
0 0
0 −2 −3 −3 1 0 0
0
3
0
0 1
0 0
R3 ( 3 ) + R2
−3 1 0
1
1
1
→
R3 ( − 2 ) + R1
0 3
fo r
R 3( 3 )→
1
( | ) 1
D is tri b
(P | I)
ut io n
→→→ P = (v1 v2 v3) =
( | ) Processing math: 100%
1
1
2
0 −3
N ot
1
0
0 −2 0 −3 1
1
0
1
0
1 1 0
1
0
1 3
0
0 1
1
R 2( − 2 )→
3
2
−3 1
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(| ) (| ) 0 0 1
1
−2 −2
2
0
0 1 0 0 0 1
−2 3 2
0
R 2( − 1) + R 1→
1
0
1
1 0 0
Solution
1
3
1
1
ut io n
0 1 0
3
−6
2
1
1
−2 −2
= (I | P − 1)
1
0
3
Hence,
D is tri b
9/9/2020
( ) 1
1
1
−6
fo r
−2 3 2
N ot
P −1 =
0
2
1
1
−2 −2 0
.
1 3
By Theorem 8.2.6,
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22/27
Solution
( )( ) 1 1 2
A 3 = PD 3P − 1 =
−2 0 0 3 0 0 0
3 1 2 0 0 3
−8 0 0
−8
0 0
0 2
= 6 − 24 0 3 0 0 3
2
2
1
1
−2 −2
0
0
1 3
−4
2
0 0 ][ 6 0 1
9 0
3
9
−1
−3 −3 ] 0 2
0
−1
−3 −3 0
3
2
24 − 24 12
1
= 6 72 − 72 54 0 0 6
fo r
4
−3
1
−3
1
8. 8.
| | λ −1 λ
N ot
12 − 12 5 . 0 0 1
=
→
( ) 3
1
−6
D is tri b
= [ 3 1 3 0 0 3
1
0 1
1
( )( ) ( ) ( )( ) ( ) ( ) 1 1 2
1. Because |λI − A | =
0
1
−2
ut io n
9/9/2020
= λ 2 + 1 = 0, the eigenvalues are λ 1 = i and λ 2 = − i.
→
For λ 1 = i, (λ 1I − A)X = 0 becomes Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
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9/9/2020
Solution
( )( ) ( ) i −1
x
1
y
i
0
=
0
.
Now we use row operations to solve the above system.
1
i
0
R 1 , 2→
( |) 1
i
0
i −1 0
R 1( − i) + R 2→
( |) 1 i 0
.
ut io n
( |) i −1 0
0 0 0
D is tri b
The system corresponding to the last augmented matrix is x + iy = 0, where x is a basic variable and y is a free variable. Let y = t. Then x = − it and
() ( ) ( ) x y
{
=t
t
}
→ tv2 : t ∈ ℝ .
2. Let D = diag(λ 1, λ 2) = diag(i − i) and
−i 1
= tv 1,
fo r
→ where v 2 = (i, 1) T. Moreover, E λ = 2
− it
=
( )
N ot
→→ −i i P = (v1 v2) = . 1 1
Then A = PDP − 1. 3. Because
Processing math: 100%
D
100
=
( ) i
0 100
0 −i
=
(
i 100
0
0
( − i) 100
)( ) =
1 0 0 1
= I2
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9/9/2020
Solution
and A 100 = PD 100P − 1 = PIP − 1 = PP − 1 = I 2.
You can check A = PDP − 1. Indeed,
( )( ) ( ) ( )( ) ( ) 1
−i i
i
1
0 −i
1
0
1
i −i
1
1
and
.
i −i
fo r
=
( )( ) ( ) 0
1
−1 0
1 −2 1
1
=
N ot AP =
1
=
1
By (1.2.5),|P | = 1 ≠ 0. By computation, we have
PB =
and
−i i
−1 0
PD =
9. 9.
1
D is tri b
0
AP =
ut io n
Remark A.8.1.
1
1
−1 2
( )( ) ( ) 1 −1 2
1
0
1
−1 0
=
1
1
−1 2
.
Hence, PB = AP and A is similar to B.
Processing math: 100%
10. 10.
Because A and B are similar, B = P − 1AP.
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9/9/2020
Solution
i. Noting that λI = P − 1(λI)P, we have |λI − B| = =
|P (λI)P − P AP | = | P (λI − A)P | |P | | λI − A | | P | = | λI − A | . −1
−1
−1
−1
− 1AP
| = |P | |A | |P | = |A | . −1
D is tri b
| | = |P
iv. B
ut io n
ii. Because |λI − A | = | λI − B | , the equations |λI − A | = 0 and |λI − B | = 0 have the same roots. Hence, A and B have the same eigenvalues. iii. Because A and B have the same eigenvalues, the result (ii) follows from Theorem 8.1.4.
v. Because B = P − 1AP and P − 1 is invertible, by Corollary 2.7.3, r(B) = r(P − 1AP) = r(AP).
r(P TA T) = r(A T) = r(A). 11. 11.
fo r
By Theorem 2.5.5 (1) r(AP) = r((AP) T). By Theorem 2.2.3 (3), r((AP) T) = r(P TA T). Because P T is invertible, by Corollaries 2.7.3 and 2.7.2,
N ot
i. Let P = I. Then P is invertible and P − 1 = I. It is obvious that A = IAI = P − 1AP and A ∼ A. ii. Because A ∼ B, there exists an invertible n × n matrix P such that B = P − 1AP. This implies that A = PBP − 1 = (P − 1)
−1
B(P − 1)
and B ∼ A. iii. Because A ∼ B and B ∼ C, there exist invertible n × n matrices P 1 and P 2 such thatB = P 1− 1AP 1 −1
and C = P 2 BP 2. Then Processing math: 100% https://etext-ise.pearson.com/print?locale=en&url=https://etext-ise.pearson.com/eps/pearson-reader/api/item/5aac7912-8886-4f9c-aa75-f326e1a3f161/1/file/9780137233267_et2_dcusb_l1/OPS/xhtml…
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9/9/2020
Solution
−1
−1
−1
−1
−1
C = P 2 BP 2 = C = P 2 (P 1 AP 1)P 2 = (P 2 P 1 )A(P 1P 2) = (P 1P 2) − 1A(P 1P 2)
Because A ∼ B, there exists an invertible P such that P − 1AP = B.
12. 12. a.
B 2 = (P − 1AP)(P − 1AP) = (P − 1A(PP − 1)AP) = (P − 1A(I)AP)
Repeating the process implies that B k = P − 1A kP.
D is tri b
= (P − 1AAP) = P − 1A 2P.
b.
ut io n
and A ∼ C.
ϕ(B) = a mB m + a m − 1B m − 1 + ⋯ + a 1B + a 0I
= a mP − 1A mP + a m − 1P − 1A m − 1P + ⋯ + a 1P − 1AP + a 0P − 1IP
N ot
fo r
= P − 1(a mA m + a m − 1A m − 1 + ⋯ + a 1A + a 0I)P = P − 1ϕ(A)P.
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A.9 Vector spaces
A.9 Vector spaces 1. Let W
= {(x, y) ∈ R
2. Let W
= {(x, y) ∈ R
2 2 +
: x − y = 0}. : x + y ≤ 1}.
Show that W is a subspace of R2 . Show that W is not a subspace of R2 .
3. Let 2
R
+
i. Show that if
→ 2 a ∈ R+
and
→ b
= {(x, y) : x ≥ 0
2
∈ R+ ,
then
and y ≥ 0}.
→ → 2 a + b ∈ R+
.
ii. Show that R2+ is not a subspace of R2 by using Definition 9.1.1
.
4. Let 2
W = R
+
+
).
→
r
→
2
∪ (−R
i. Show that if a ∈ W and k is a real number, then k a ∈ W . ii. Show that W is not a subspace of R2 by using definition 9.1.1 5. Let W 9.1.1
3
= {(x, y, z) ∈ R
and Theorem 9.1.1
: x − 2y − z = 0}.
.
Show that W is a subspace of R3 by Definition
, respectively.
6. Show that W is a subspace of R4 , where 4
W = {(x, y, z, w) ∈ R
: {
x − 2y + 2z − w = 0, −x + 3y + 4z + 2w = 0
7. Show that W is a subspace of R3 , where
…
1/2
}.
9/9/2020
A.9 Vector spaces x + 2y + 3z = 0, ⎧ ⎪
⎧ ⎪ 3
W = ⎨(x, y, z) ∈ R ⎩ ⎪
8. Let W
3
= {(x, y, z) ∈ R
: x − 2y − z = 3}.
: ⎨ −x + y − 4z = 0, ⎬ . ⎩ ⎭ ⎪ ⎪ x − 2y + 5z = 0
Show that W is not a subspace of R3 by Definition
9.1.1 . 9. Let W = [0, 1]. Show that W is not a subspace of R. 10. Show that all the subspaces of R are {0} and R . 11. Let W = {(0, 0, ⋯ , 0)} ∈ Rn . Show that W is a subspace of Rn . 12. Let
→ n υ ∈ R
be a given nonzero vector. Show that → W = {t υ : t ∈ R}
is a subspace of Rn by Definition 9.1.1
…
2/2
.
⎫ ⎪
9/9/2020
A.9 Vector spaces
A.9 Vector spaces 1. Let W
= {(x, y) ∈ R
2. Let W
= {(x, y) ∈ R
2 2 +
: x − y = 0}. : x + y ≤ 1}.
Show that W is a subspace of R2 . Show that W is not a subspace of R2 .
3. Let 2
R
+
i. Show that if
→ 2 a ∈ R+
and
→ b
= {(x, y) : x ≥ 0
2
∈ R+ ,
then
and y ≥ 0}.
→ → 2 a + b ∈ R+
.
ii. Show that R2+ is not a subspace of R2 by using Definition 9.1.1
.
4. Let 2
W = R
+
+
).
→
r
→
2
∪ (−R
i. Show that if a ∈ W and k is a real number, then k a ∈ W . ii. Show that W is not a subspace of R2 by using definition 9.1.1 5. Let W 9.1.1
3
= {(x, y, z) ∈ R
and Theorem 9.1.1
: x − 2y − z = 0}.
.
Show that W is a subspace of R3 by Definition
, respectively.
6. Show that W is a subspace of R4 , where 4
W = {(x, y, z, w) ∈ R
: {
x − 2y + 2z − w = 0, −x + 3y + 4z + 2w = 0
7. Show that W is a subspace of R3 , where
…
1/2
}.
9/9/2020
A.9 Vector spaces x + 2y + 3z = 0, ⎧ ⎪
⎧ ⎪ 3
W = ⎨(x, y, z) ∈ R ⎩ ⎪
8. Let W
3
= {(x, y, z) ∈ R
: x − 2y − z = 3}.
: ⎨ −x + y − 4z = 0, ⎬ . ⎩ ⎭ ⎪ ⎪ x − 2y + 5z = 0
Show that W is not a subspace of R3 by Definition
9.1.1 . 9. Let W = [0, 1]. Show that W is not a subspace of R. 10. Show that all the subspaces of R are {0} and R . 11. Let W = {(0, 0, ⋯ , 0)} ∈ Rn . Show that W is a subspace of Rn . 12. Let
→ n υ ∈ R
be a given nonzero vector. Show that → W = {t υ : t ∈ R}
is a subspace of Rn by Definition 9.1.1
…
2/2
.
⎫ ⎪
9/9/2020
Solution
Solution →
1. 1.
a = (x 1 , y 1) ∈ W if and only if x 1 − y 1 = 0. Let → a = (x 1, y 1) ∈ W, b = (x 2, y 2) ∈ W, and Note that →
k ∈ ℝ is a real number. Then x 1 − y 1 = 0 and x 2 − y 2 = 0, →
→
a + b = (x 1, y 1) + (x 2, y 2) = (x 1 + x 2, y 1 + y 2) and k→ a = (kx 1, ky 1).
Because (x 1 + x 2) − (y 1 + y 2) = (x 1 − y 1) + (x 2 − y 2) = 0 + 0 = 0, and kx 1 − ky 1 = k(x 1 − y 1) = k(0) = 0, we have →
→
a + b = (x 1 + x 2, y 1 + y 2) ∈ W
and k→ a = (kx 1, ky 1) ∈ W. Hence, W is a subspace of ℝ 2. 2. 2. Let (x, y) = (0, 1) and k = 2. Then k(x, y) = 2(0, 1) = (0, 2) and kx + ky = 0 + 2 > 1. This implies that k(x, y) ∈ W. 3. 3. 2
2
→
i. Let → a = (x1, y1) ∈ ℝ + and b = (x 2, y 2) ∈ ℝ + . Then x 1 ≥ 0, y 1 ≥ 0, x 2 ≥ 0, and y 2 ≥ 0. Then a+→ a = (x 1, y 1) + (x 2, y 2) = (x 1 + x 2, y 1 + y 2) ∈ ℝ 2+
→
because x 1 + x 2 ≥ 0 and y 1 + y 2 ≥ 0. 2
2
ii. Let → a = (1, 1) and k = − 1, then → a ∈ ℝ + and k→ a = ( − 1, − 1) ∈ W. Hence, ℝ + is not a Processing math: 100% …
1/6
subspace of ℝ 2.
9/9/2020
Solution
4. 4. Note that W is the union of the first and third quadrants. Also, (x, y) ∈ W if and only if either x ≥ 0 and y ≥ 0 or x ≤ 0 and y ≤ 0. i. Let → a = (x, y) ∈ W. Then either x ≥ 0 and y ≥ 0 or x ≤ 0 and y ≤ 0. This implies that → k a = (kx, ky) satisfies either kx ≥ 0 and ky ≥ 0 or kx ≤ 0 and ky ≤ 0. Hence, k→ a ∈ W. →
→
ii. Let → a = (1, 2) and b = ( − 2, − 1), then → a ∈ W and b ∈ W. But →
a + b = (1, 2) + ( − 2, − 1) = ( − 1, 1) ∈ W.
→
Hence, W is not a subspace of ℝ 2. 5. 5.
→
Let → a = (x 1, y 1, z 1) ∈ W, b = (x 2, y 2, z 2) ∈ W, and k ∈ ℝ is a real number. Then x 1 − 2y 1 − z 1 = 0 and x 2 − 2y 2 − z 2 = 0, →
a + b = (x 1, y 1, z 1) + (x 2, y 2, z 2) = (x 1 + x 2, y 1 + y 2, z 1 + z 2).
→
and k→ a = (kx 1, ky 1, kz 1). Because (x 1 + x 2) − 2(y 1 + y 2) − (z 1 + z 2) = (x 1 − 2y 1 − z 1) + (x 2 − 2y 2 − z 2) = 0 + 0 = 0, and kx 1 − 2ky 1 − kz 1 = k(x 1 − 2y 1 − z 1) = k(0) = 0, we have →
a+b∈W
→
→
and k→ a ∈ W. Hence, by Definition 9.1.1, W is a subspace of ℝ 3. Let A = 1 − 2 − 1 , X = (x, y, z) T, and →
0 = (0, 0, 0) T.
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9/9/2020
Solution
W = =
{(x, y, z) ∈ ℝ : x − 2y − z = 0 } {(x, y, z) ∈ ℝ : AX = 0 }. 3
→
3
→
By Theorem 9.1.1, W is a subspace of ℝ 3. 6. 6.
Let A =
(
1 −1
−2 2 −1 3
4
2
)
→
→
, X = (x, y, z) T, and 0 = (0, 0, 0) T. Then
W = =
{
(x, y, z, w) ∈ ℝ 4 :
{
x − 2y + 2z − w = 0, − x + 3y + 4z + 2w = 0
{(x, y, z, w) ∈ ℝ : AX = 0 }. →
4
→
By Theorem 9.1.1, W is a subspace of ℝ 4.
7. 7.
Let A =
(
1 −1 3
2
3
)
→ → −4 , X = (x, y, z) T, and 0 = (0, 0, 0) T. Then −2 5
1
W =
= Theorem Processing math: By 100% …
3/6
{
(x, y, z) ∈ ℝ 3 :
{
x + 2y + 3z = 0, − x + y − 4z = 0, x − 2y + 5z = 0
{(x, y, z) ∈ ℝ : AX = 0 }.
9.1.1, W is a subspace of ℝ 3.
3
→
→
}
}
9/9/2020
Solution
8. 8.
Note that → a = (x 1, y 1, z 1) ∈ W if and only if x 1 − 2y 1 − z 1 = 3. Because 0 + 2(0) + 3(0) ≠ 3, so
(0, 0, 0) ∈ W and W is not a subspace of ℝ 3. 9. 9. Let x = 1 and k = 2. Then x ∈ [0, 1] = W and kx = 2(1) = 2 > 1. This implies that kx ∈ W and W is not a subspace of R. 10. 10. It is easy to show that {0} and ℝ are subspaces of R. Let W be a subset of R satisfying W ≠ {0} and W ≠ ℝ. Then there exist two points x 1 satisfying x 1 ∈ W and x 1 ≠ 0 and x 2 ∈ ℝ satisfying x 2 ∈ W. Let k = 11. 11.
x2 x1
. Then x 1 ∈ W but kx 1 = x 2 ∈ W. Hence, W is not a subspace of R.
Note that W only contains the origin (0, 0, ⋯, 0), so a = (x 1, x 2, ⋯, x n) ∈ W if and only if x 1 = x 1 = ⋯ = x n = 0.
→
→
Let → a = (x 1, x 2, ⋯, x n) ∈ W, b = (y 1, y 2, ⋯, y n) ∈ W, and k ∈ ℝ is a real number. Then x i = y i = 0 for i = 1, 2, ⋯, n. It follows that →
a + b = (x 1, x 2, ⋯, x n) + (y 1, y 2, ⋯, y n)
→
= (x 1 + y 1, x 2 + y 2, ⋯, x n + y n) = (0 + 0, 0 + 0, ⋯, 0 + 0) = (0, 0, ⋯, 0) ∈ W. and k→ a = (kx 1, kx 2, ⋯, kx n) = (k(0), k(0), ⋯, k(0)) = (0, 0, ⋯, 0) ∈ W. Hence, W is a subspace of ℝ n. 12. 12.
→
Note that Note that → a ∈ W if and only if there exists t ∈ ℝ such that → a = t→ v. Let → a ∈ W, b ∈ W, →
and k ∈ ℝ. Then there exist t 1, t 2 ∈ ℝ such that → a = t 1→ v and b = t 2→ v. Hence, Processing math: 100% …
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→
→
a + b = t 1→ v = (t 1 + t 2)→ v ∈W
9/9/2020
Solution
and k→ a = k(t 1→ v) = (kt 1)→ v ∈ W.
Section 8.3
n
1. Let
tri
bu
Show that C(ℝ) is a vector space under the following operations. 1. + : (f + g)(x) = f(x) + g(x) for x ∈ ℝ. 2. (kf)(x) = k(f(x)) for x ∈ ℝ.
tio
C(ℝ) = {f : f : ℝ → ℝ is continuous on ℝ}.
is
2. Let
rD
C[a, b] = {f : f : [a, b] → ℝ is continuous}.
3. Let n be a nonnegative integer and let
{p : p(x) = a
N
Pn =
ot
fo
Show that C[a, b] is a vector space under standard addition and scalar multiplication: 1. For f, g ∈ C[a, b], (f + g)(x) = f(x) + g(x) for x ∈ [a, b]. 2. For k ∈ ℝ and f ∈ C[a, b], (k ⋅ f)(x) = kf(x) for x ∈ [a, b]
0
Show that P n is a subspace of C(ℝ). 4. Let Processing math: 100% …
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}
+ a 1x + ⋯ + a nx n is a polynomial of degree n .
V = {A : A is a 3 × 3 matrix}.
9/9/2020
Solution
We define the following addition and scalar multiplication. 1. Addition: For A, B ∈ V, A + B = AB, where AB is the standard product of A and B in Section 2.5 . 2. Scalar multiplication: For A ∈ V and k ∈ ℝ, k ⋅ A = kA, where kA is the standard scalar multiplication given in Definition 2.1.4 . Then V is not a vector space under the above addition and scalar multiplication. 5. Let C 1[a, b] =
{f : f
′
}
∈ C[a, b] ,
where f ′ (x) denotes the derivative of f at x. Show that C 1[a, b] is a vector subspace of C[a, b].
Processing math: 100% …
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9/9/2020
Solution
Solution If f, g ∈ C(R), then f + g ∈ C(R), and if k ∈ R and f ∈ C(R), then kf ∈ C(R). Hence, 1. 1. the axioms (1) and (6) hold. It is easy to verify that the other eight axioms given in Definition 9.2.2 hold. Hence, C(R) is a vector space. If f, g ∈ C[a, b], then f + g ∈ C[a, b], and if k ∈ R and f ∈ C[a, b], than kf ∈ C[a, b]. 2. 2. Hence, the axioms (1) and (6) hold. It is easy to verify that the other eight axioms given in Definition 9.2.2 hold. Hence, C[a, b] is a vector space. For the set V, the axioms (2) and (4) are not true in general. For example, let 3. 3. 1
0
A = ⎜0
0
⎛
⎝
0
⎞ 0⎟
0
0
⎠
0
0
B = ⎜0
0
0⎟.
0
0
⎛
⎝
0
⎞
⎠
Then A + B = AB = B ≠ A, so the axiom does not hold. Also, Let 0
0
2
A = ⎜0
0
0⎟
0
0
⎛
⎝
0
⎞
⎠
0
2
1
and B = ⎜ 0
0
0⎟.
0
0
⎛
⎝
0
⎞
⎠
Then 0
2
1
AB = ⎜ 0
0
0⎟
⎛
⎝
0
0
0
⎞
⎠
0
2
4
and BA = ⎜ 0
0
0⎟
0
0
⎛
⎝
0
⎞
⎠
So A + B = AB ≠ BA = B + A and axiom (2) does not hold. It is easy to verify that other axioms hold. Hence, V is not a vector space under the above addition (1) and scalar multiplication (2). …
1/2
9/9/2020
Solution
4. 4. Pn
Let p ∈ Pn . Because p is polynomial, p is continuous on R and thus, p ∈ C(R). Hence, ⊂ C(R). Let p, q ∈ P n and k ∈ R. Then p + q ∈ P n and kp ∈ P n . Hence, P n is a subspace of
C(R).
Let f, g ∈ C 1 [a, b] and k ∈ R, then f ′ , g ′ ∈ C[a, b]. This implies ′ ′ ′ 1 1 (f + g) = f + g ∈ C [a, b]. Hence, C [a, b] is a vector subspace of C[a, b].
5. 5.
…
2/2
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A.10 Complex numbers
A.10 Complex numbers Section 10.1 1. Solve each of the following equations. a. z 2 + z + 1 = 0 b. z 2 − 2z + 5 = 0 c. z 2 − 8z + 25 = 0 2. Use the following information to find x and y. a. 2x − 4i = 4 + 2yi b. (x − y) + (x + y)i = 2 − 4i 3. Let z1 = 2 + i and z2 = 1 − i. Express each of the following complex numbers in the form x + yi, calculate its modulus, and find its conjugate. 1. z1 + 2z2 2. 2z1 − 3z2 3. z1 z2 z 4. z 1 2
4. Express each of the following complex numbers in the form x + yi . a. b. c. d.
…
1/2
¯¯¯¯ ¯¯¯ ¯ i ( ) 1−i
2−i (1−i)(2+i) 1+i i(1+i)(2−i) 1+i 1−i
−
2−i 1+i
9/9/2020
A.10 Complex numbers
5. Find i2 , i3 , i4 , i5 , i6 and compute (−i − i4 1−i
8
6. Let z = ( 1+i )
.
Calculate z 66
+ 2z
33
+i
5
6
−i )
1000
.
− 2.
7. Find z ∈ C a. iz = 4 + 3i b. (1 + i)z¯ = (1 − i)z 8. Find real numbers x, y such that x + 1 + i(y − 3) = 1 + i. 5 + 3i
9. Let z =
(3+4i)(1+i) 5
i (2+4i)
…
2/2
2
12
.
Find |z|.
9/9/2020
Solution
Solution 1. 1. a. z 1 = b. z 1 = c. z 1 =
− 1 + i √3 2 2 + i√16 2 8 + i√36 2
and z 2 =
− 1 + i √3 2
= 1 + 2i and z 2 = = 4 + 3i and z 2 =
. 2 + i√16 2 8 + i√36 2
= 1 − 2i. = 4 − 3i.
2. 2. a. By Definition 10.1.2, we have 2x = 4 and − 4 = 2y. This implies that x = 2 and y = − 2. b. By Definition 10.1.2, we have x − y = 2 and x + y = − 4. Solving the system, we obtain x = − 1 and y = − 3. 3. 3. z 1 + 2z 2 = (2 + i) + 2(1 − i) = (2 + i) + (2 − 2i) = 4 − i, 1.
|z1 + 2z2 | = |4 − i| = √42 + ( − 1)2 = √17, ¯ ¯ and z 1 + 2z 2 = 4 − i = 4 + i. 2z 1 − 3z 2 = 2(2 + i) − 3(1 − i) = (4 + 2i) − (3 − 3i) = 1 + 5i,
2.
|2z1 − 3z2 | = |1 + 5i| = √12 + 52 = √26 ¯ ¯ and 2z 1 − 3z 2 = 1 + 5i = 1 − 5i.
…
1/5
9/9/2020
Solution
z 1z 2 = (2 + i)(1 − i) = 2(1 − i) + i(1 − i) = 2 − 2i + i − i 2 = 2 − i + 1 = 3 − i,
| |
3. z z 1 2
= |3 − i| =
√32 + ( − 1) 2 = √10
and
¯ ¯ z 1z 2 = 3 − i = 3 + i. z1
4.
2+i
=
z2
1−i
||
1+1
2(1+i) +i(1+i)
=
1 − i2
2 + 3i − 1
=
=
2
1 2
3
+ 2 i,
| | √( ) ( ) 1
=
z2
(1−i) (1+i)
2 + 2i + i + i 2
= z1
(2+i) (1+i)
=
1 2
3
+ 2i =
2
¯
2
3 2
+
2
=
√10 2
¯
() z1
( ) 1
=
z2
2
3
+ 2i =
1 2
3
− 2 i.
4. 4. a.
i 1−i
=
i(1+i) (1−i) (1+i)
¯ 1−i
…
2/5
i−1 2
= −
1 2
1
1
)
= − 2 + 2i = −
1 2
1
1
+ 2 i,
¯
( ) ( i
=
− 2 i.
and
9/9/2020
Solution
2−i
b.
(1−i) (2+i)
= =
1+i
c.
i(1+i) (2−i)
=
(2−i) 2(1+i) (1−i) (1+i) (2+i) (2−i) 7−i 10
d.
1−i
= 5. 5.
−
2i 2
2−i 1+i
−
=
1 − 3i 2
7
−
1
4i − 2
10 (1+i) 2
(1−i) (1+i)
=i−
1 2
1
=
5
−
( 3 − 4i ) ( 1 + i ) (2) (5)
i.
10 10 i(1+i) (1−i) (2+i)
i2 ( 1 + i ) ( 1 − i ) ( 2 − i ) ( 2 + i )
= − 1+i
=
=
=
2i ( 2 + i ) − (2) (5)
2
− 5 i. (2−i) (1−i) (1+i) (1−i)
3
+ 2i = −
1 2
5
+ 2 i.
Because i 2 = − 1, i 3 = i 2 ⋅ i = − i, i 4 = i 2 ⋅ i 2 = ( − 1)( − 1) = 1, i 5 = i 4 ⋅ i = i, and
i 6 = i 5 ⋅ i = i ⋅ i = − 1, − i − i 4 + i 5 − i 6 = − i − 1 + i − ( − 1) = 0 and ( − i − i 4 + i 5 − i 6) 6. 6. z=
Because
( ) 1−i
1+i
8
1−i 1+i
=
(1−i) 2 (1+i) (1−i)
=
1 − 2i − 1 2
1000
= 0 1000 = 0.
= − i,
4
= ( − i) 8 = [( − i) 2] = ( − 1) 4 = 1
and z 66 + 2z 33 − 2 = 1 + 2 − 2 = 1. 7. 7. a. Because iz = 4 + 3i, i(iz) = i(4 + 3i). This implies that i 2z = 4i + 3i 2 and − z = 4i − 3. Hence, z = 3 − 4i. b. Let z = x + yi. Then zˉ = x − yi and …
3/5
9/9/2020
Solution
(1 + i)(x − yi) = (1 − i)(x + yi). Hence, x − yi + xi + y = x + yi − xi + y and 2(x − y)i = 0. This implies x = y and z = x(1 + i) for x ∈ ℝ. 8. 8.
Because x+1+i(y−3) 5 + 3i
= = =
[ x + 1 + i ( y − 3 ) ] ( 5 − 3i ) ( 5 + 3i ) ( 5 − 3i ) [ ( x + 1 ) ( 5 − 3i ) + i ( y − 3 ) ( 5 − 3i ) 25 + 9 ( 5x + 3y − 4 ) + i [ − 3 ( x − 1 ) + 5 ( y − 3 ) ] 34
= i + 1, comparing the real and imaginary parts implies that
{
5x + 3y − 4 34
=1
−3(x−1) +5(y−3) 34
= 1.
Simplifying the above equation, we obtain
{
5x + 3y = 38 − 3x + 5y = 52.
Solving the above equation, we have x = 1 and y = 11.
…
4/5
9/9/2020
Solution
|
| 3 + 4i | ( 1 + i ) 12
9. 9.
|z| =
| i | | ( 2 + 4i ) | 5
2
|
=
| 3 + 4i | | 1 + i | 12 | i | 5 | 2 + 4i | 2
=
5 ( √2 )
12
( 1 ) 5 ( 20 )
=
26 4
= 16.
Section 10.2 1. For each of the following numbers, find its principal argument and argument. a. − 4 b. − i c.
1 2
−
√3 2
i
d. 1 + √3i e. − 1 + √3i f. − √3 − i 2. Express each of the following complex numbers in polar and exponent forms. a. 1 b. i c. − 2 d. − 3i e. 1 + √3i f. − 1 − i
…
5/5
9/9/2020
Solution
Solution 1. 1.
By Theorem 10.2.1, we have a. Arg( − 4) = π and arg( − 4) = π + 2kπ for k ∈ ℤ. b. Arg( − i) = −
π 2
c. Let z = x + yi =
and arg( − i) = − 1 2
−
√3 2
π 2
i. Then x =
+ 2kπ for k ∈ ℤ. 1 2
√3
and y = −
2
. Let ψ ∈ (0,
π 2
)
satisfy
tanψ =
|| y
x
=
√3.
π
Then ψ = 3 . Because x > 0 and y < 0, z is in the fourth quadrant and by Theorem 10.2.2 (4), Arg z = − ψ = −
π 3
d. Let z = x + yi = 1 + √3i. Then x = 1 and y =
and arg z = −
√3.
tanψ =
π 3
Let ψ ∈ (0,
|| y
x
=
+ 2kπ for k ∈ ℤ. π 2
) satisfy
√3.
π
Then ψ = 3 . Because x > 0 and y > 0, z is in the first quadrant and by Theorem 10.2.2 (1), Arg z = ψ = …
1/4
π 3
and arg z =
π 3
+ 2kπ k ∈ ℤ.
Solution
e. Let z = x + yi = − 1 + √3i. Then x = − 1 and y = tanψ =
√3. Let ψ ∈ (0,
|| y
x
=
π 2
) satisfy
√3.
π
Then ψ = 3 . Because x < 0 and y > 0, z is in the second quadrant and by Theorem 10.2.2 (2), Arg z = π − ψ = π −
π 3
=
2π 3
and arg z =
2π 3
+ 2kπ for k ∈ ℤ.
f. Let z = x + yi = − √3 − i. Then x = − √3 and y = − 1. Let ψ ∈ (0, tanψ =
|| y
x
=
√3 3
π 2
) satisfy
.
π
Then ψ = 6 . Because x < 0 and y < 0, z is in the third quadrant and by Theorem 10.2.2 (3), Arg z = − π + ψ = − π +
and arg z = −
5π 6
+ 2kπ for k ∈ ℤ.
2. 2. a. 1 = 1(cos 0 + i sin 0) = e 0 . i.
(
π
b. i = 1 cos 2 + i sin
π 2
)
π
= e 2 i.
c. − 2 = 2(cosπ + i sinπ) = 2e πi. …
2/4
π 6
= −
5π 6
Solution
( ( ) ( )) π
d. − 3i = 3 cos − 2 e. r = |z| =
√
π
+ isin − 2
x2 + y2 =
π
= 3e
− 2i
.
√1 + 3 = 2. Let z = x + yi = 1 + √3i.
Then x = 1 and y =
√3.
( )
Let ψ ∈ 0,
satisfy
tanψ =
|| y
x
=
√3 1
=
√3.
π
Then ψ = 3 . Because x > 0 and y < 0, z is in the first quadrant and by Theorem 10.2.2 (1), π
Arg z = ψ = 3 . Hence,
(
1 + √3i = 2 cos
f. r = |z| =
√( − 1) 2 + ( − 1) 2 = √2.
( )
ψ ∈ 0,
π
2
π 3
+ i sin
π 3
)
π
= 2e i 3 .
Let z = x + yi = − 1 − i. Then x = − 1 and y = − 1. Let
satisfy
tanψ =
| | −1
−1
= 1.
π
Then ψ = 4 . Because x < 0 and y < 0, z is in the third quadrant and by Theorem 10.2.2 (3), Arg z = − π + ψ = − π + …
3/4
π 4
= −
3 4
π.
π
2
Solution
Hence,
−1 − i =
√2
( ( ) ( )) √ ( cos −
3π
+ i sin −
4
3π 4
=
2 cos
3π 4
Section 10.3 1. Express the following numbers in x + yi form. 1. (1 + √3i)
3
2. ( − 1 − i) 4 3. (1 + i) − 8. 2. Let z 1 = 3. Let z =
1+i
√2 √2 2
and z 2 =
√3 − i.
Find the exponential form of z 1z 2.
(1 − i). Find z 100 + z 50 + 1.
4. Compute the following value
(
z= −
…
4/4
1 2
+
√3 2
) ( 600
i
+ −
1 2
−
√3 2
i
)
60
.
− i sin
3π 4
)
3π
=
e −i 4
.
Solution
Solution 1. 1.
(
π
)
π
1. Because 1 + √3i = 2 cos 3 + i sin 3 ,
(1 + √3i)
(
3
) (
π 3
π
= 2 3 cos 3 + i sin 3
= 8 cos3 ⋅
π 3
+ i sin3 ⋅
π 3
= 8(cosπ + i sinπ) = 8( − 1 + 0) = − 8.
2. Because ( − 1 − i) =
(
√2 cos
3π
− i sin
4
3π 4
)
,
( − 1 − i) 4 = 4(cos3π − i sin3π) = − 4.
3. Because 1 + i =
√2
(
π
π
cos 4 + i sin 4
(1 + i)
−8
=
)
π
= √2 e i 4 ,
( ) i
π
√2e 4
−8
=
( √2 )
= 2 − 4e − i2π = 2 − 4 2. 2.
…
1/4
Because
−8
( ) e
i
π
4
−8
=2
−4 i
e
− 8π 4
)
Solution
z1 = z2 =
1+i
√2
π
π
π
= cos 4 = + i sin 4 = e i 4 and
(
π
π
√3 − i = 2 cos 6 − i sin 6
)
π
= 2e i 6 ,
we have π
π
z 1z 2 = 2 e i 4 e i 6 = 2e i
3. 3.
Because z =
√2 2
z
( ) π π + 4 6
5π
= 2e i 12 .
π
(1 − i) = e − 4 ,
100
( ) π
=
e
−i4
100
100π
=
e −i 4
= e 2 ( − 12 ) πi + ( − π ) i = e 2 ( − 12 ) πie ( − π ) i
= e ( − π ) i = cos( − π) + i sin( − π) = cosπ = − 1 and
z 50 =
( ) π
e −i4
50
π
( ) ( ) π
= cos − 2
π
π
+ i sin − 2
( ) π
= i sin − 2
= − i.
Hence, z 100 + z 50 + 1 = − 1 − i + 1 = − i. …
2/4
π
= e 2 ( − 6 ) πi ( − 2 ) i = e 2 ( − 6 ) πie ( − 2 ) i = e ( − 2 ) i
Solution
4. 4.
Let z 1 = −
1 2
+
√3 2
i and z 2 = −
1
−
2
√3 2
i. Then
( ) ( )
z 1 = cos π −
π 3
+ i sin π −
π 3
= cos
2π 3
+ i sin
2π 3
2π
= ei 3
and
(
z 2 = cos − π +
π 3
) (
+ i sin − π +
π 3
) ( ) ( ) = cos −
2π 3
+ i sin −
2π 3
2π
=
e −i 3
Hence,
600 z1
600
This implies that z = z 1
( ) 2π
=
ei 3
600
=e
i400π
= 1 and
60 z2
60
+ z 2 = 1 + 1 = 2.
Section 10.4 1. Find all roots for each of the following complex numbers. a.
4
√1 6 b. √− 1 c.
3
√1 − i 3 d. √− i 1
e. ( − 1 + √ …
3/4
3i) 4
( ) 2π
=
e −i 3
60
= e − i40 = 1.
.
Solution
2. Find all solutions of z 3 = − 8. 3. Solve the following equations. a. (1 − i)z 2 + 2z + (1 + i) = 0 b. z 2 + 2z + (1 − i) = 0
…
4/4
Solution
Solution 1. 1.
(
4
√1 cos
= cos
kπ 2
0 + 2kπ 4
+ i sin
kπ 2
+ i sin
0 + 2kπ 4
)
= cos
4
π
4
4
rD
4
}
6
√− 1 = It follows that
…
1/6
1
N
b. Because − 1 = e iπ, we have
6π 4
+ i sin
6π 4
ot
1, i, − 1, − i .
(e iπ) 6
=e
4
π
fo
{
2kπ
( √1) 1 = cos 2 + i sin 2 = i,
( √1) 2 = cosπ + i sinπ = − 1, and ( √1) 3 = cos
√1 =
4
is
( √1) 0 = cos0 + i sin0 = 1,
4
+ i sin
for k = 0, 1, 2, 3.
Hence, we obtain
Hence,
2kπ
tri bu tio
4
( √1) k =
π + 2kπ 6
n
a. Because 1 = 1 ⋅ (cos0 + i sin0),
for k = 0, 1, 2, 3, 4, 5.
= − i.
Solution
( √− 1) 0 = e 6 =
2
5π
6
( √− 1) 2 = e
6
= −
1
√3
( √− 1) 1 = e 2 = i, 1
+ 2 i, ( √− 1) 3 = e
2
= e
( √− 1) 4
2
6
= −
and ( √− 1) 5 = e
6
2
√3
11π
6
= −i
√3
7π
6
3π
6
π
6
+ 2i
=
2
1
− 2 i.
Hence,
√1 − i = √2e
−4
1 3
=
6
√2e
3
( √1 − i) 0 = 3
3
( √1 − i) 2 =
…
2/6
3
√1 − i =
{
3
6
−
π 4
6
+ 2kπ
for k = 0, 1, 2,
3
−
6
√2e 6
π 4
=
3
6
3
6
(
π
−
√2e
π
4
+ 2π
3
−
6
√2e
3
π
4
=
+ 2π
3
π
)
√2 cos 12 + i sin 12 ,
N ot
( √1 − i) 1 =
Hence,
6
}
rib
( ) π
3
6
is t
c. Because
{
6
( √− 1) 0, ( √− 1) 1, ( √− 1) 2, ( √− 1) 3, ( √− 1) 4, ( √− 1) 5 .
D
√− 1 =
fo r
6
1
− 2 i,
ut io n
√3
π
6
=
}
( √ ( 6
7π
7π
) )
√2 cos 12 + i sin 12 , . 6
( √1 − i) 0, ( √1 − i) 1, ( √1 − i) 2 .
2 cos
5π 4
+ i sin
5π 4
Solution
d. Because
( ) π
3
√1 − i =
e −i2
( )
1 3
−
=
e
π
+ 2kπ
2
for k = 0, 1, 2,
3
π
3
π
π
( √− i) 0 = e − i 6 = cos 6 − i sin 6 = 3
π
3
ei 6
π
√3 2
1
− i2,
π
( √− i) 1 = e i 2 = cos 2 + i sin 2 = i, 7π
( √− i) 2 =
Hence,
3
√− i =
{
√3 2
1
− i 2 , i, −
√3 2
1
6
+ isin
7π 6
= −
√3 2
}
π
π
) (
− 1 + √3i = 2 cos(π − 3 ) + i sin(π − 3 ) = 2 cos
1
( − 1 + √3i) 4 =
For k − 0, 1, 2, 3, we obtain
…
3/6
1
− i2.
− i2 .
(
e.
= cos
7π
4
(
√2 cos
2π 3
2π
+ 2kπ 4
+ i sin
3
+ 2kπ 4
)
2π 3
+ i sin
2π 3
)
,
for k = 0, 1, 2, 3.
Solution
( − 1) + √3i) 4
[
]
[ [ [
] ] ]
1
1
(−1+√
3i) 4
1
(−1+√
3i) 4
1
( − 1 + √3i) 4
1
Hence,( − 1 + √
2. 2.
3i) 4
=
2
3
=
{[
4
√2
=
=
( − 1) + √3i) 4
π
√2 cos 6 + i sin 6
=
1
[
√− 8 = 2 cos( Hence,
4/6
1
(
π
4
(
cos
(
√2 cos 4
(
√2 cos
]
k
2π 3
7π 6
5π 3
+ i sin
+ i sin
+ i sin
) ( =
2π 3
7π 6
5π 3
4
√2
√3 2
) √( =
4
1
4
√2 −
) ( =
4
π + 2kπ
√2
1 2
√3 2
−i
}
: k = 0, 1, 2, 3 .
3
) + i sin(
π + 2kπ 3
]
)
2 −2 +i
) ( =
1
+ i2 ,
Because − 8 = 8(cosπ + i sinπ), we have 3
…
0
4
) for k = 0, 1, 2.
√3 2
1
) )
,
− i2 ,
√3 2
)
.
Solution
π
π
( √− 8) 0 = 2 cos 3 + sin 3 3
( √− 8) 1 = 2 cos( 3
( √− 8) 2 = 2 cos
π + 2π
5π 3
3
) ( =2
) + i sin(
+ i sin
5π 3
1 2
√3
+
π + 2π 3
) ( =2
2
)
i = 1 + √3i,
)
) = 2(cosπ + i sinπ) = − 2,
1 2
−
√3 2
)
i = 1 − √3i.
3
All solutions of z 3 = − 8 are ( √− 8) k for k = 0, 1, 2.
rib
3. 3.
2(1 − i)
−1+i
2(1 − i)
1−i
=
− 1 + √− 1
= − 1 and
fo r
By Example 10.4.1, √− 1 = ± i. Hence, z 0 =
=
− 2 + √− 4
D
− 2 + √4 − 4(1 − i)(1 + i)
is t
a. By (10.4.9), the solution of the equation is
z=
N ot
(1 + i) 2 2i z1 = − = − = − i. (1 − i)(1 + i) 2
Hence, z 0 = − 1 and z 1 = − i are the solutions of (1 − i)z 2 + 2z + (1 + i) = 0. b. By (10.4.9), the solution of the equation is
…
5/6
ut io n
( ( (
3
(1 − i)
.
Solution
z=
2 (e i 2 )
(
=
ei
π 2
= − 1 + √i
+ 2kπ ) 2
for k = 0, 1, π
= − 1 + cos
and 5π
z1 = − 1 +
ei 4
π 4
+ i sin
π 4
fo ot N 6/6
2
+i
√2 2
√2 √2 5π 5π = − 1 + cos + i sin = −1− −i . 4 4 2 2
Hence, z 0 and z 1 are the solutions of z 2 + 2z + (1 − i) = 0.
…
= −1+
√2
rD is tri bu t
z0 = − 1 + e
i4
n
Because √i =
1
2
io
π
− 2 + √4 − 4(1 − i)