A portrait of linear algebra 3rd edition pdf

Linear Algebra

Chapter Zero Exercises

1. A True logical statement. 2. A logical statement, but it is False, because −5  3 but 25  9. 3. A True logical statement, using the properties of inequalities found in Appendix A. 4. A False logical statement, because if x  0, then x is imaginary. 5. A True logical statement as of June 2009, with 237 consecutive weeks. 6. Not a logical statement, because it cannot be ascertained to be True or False (“best” is not

a well-defined adjective; unlike the previous Exercise, where “most number of consecutive weeks as number 1” is well defined).

7. Converse: If you can watch TV tonight, then you did your homework before dinner. Inverse: If you do not do your homework before dinner, you cannot watch TV tonight. Contrapositive: If you cannot watch TV tonight, then you did not do your homework before dinner.

8. Converse: If we don’t go to the beach tomorrow, then it rained. Inverse: If it doesn’t rain tomorrow, we will go to the beach. Contrapositive: If we go to the beach tomorrow, then it did not rain.

9. Converse: If cosx ≥ 0, then 0 ≤ x ≤ /2. Inverse: If x  /2 or x  0, then cosx  0. Contrapositive: If cosx  0, then x  /2 or x  0.

10. If fx is continuous on the closed interval a, b then it possesses both a maximum and a minimum on a, b. Converse: If fx possesses both a maximum and a minimum on a, b, then fx is continuous on a, b. Inverse: If fx is not continuous on a, b, then fx either does not possess an absolute maximum or an absolute minimum on a, b. Contrapositive: If fx does not possess either an absolute maximum or an absolute minimum on a, b, then fx is not continuous at x  a.

11. A  B  a, b, c, f, g, h, i, j, m, p, q, A ∩ B  c, h, j, A − B  a, f, i, m, B − A  b, g, p, q.

12. A  B  a, b, d, g, h, j, k, p, q, r, s, t, v, A ∩ B  d, g, h, p, t, A − B  a, j, r, B − A  b, k, q, s, v.

23. If there were a largest positive number x, what can you say about x  1? 27. “If n does not have a prime factor which is at most n , then n is prime.” The number

11303 is composite. One prime factor is smaller than 100. 38. 2027 and 2029. 39. 233 49. Hint: In Step 3, write 2n1 as 22n  2n  2n. 54. f. For any two sets X and Y : X ∩ Y ⊆ X and X ∩ Y ⊆ Y. 55. a.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29 58. a. ∅, a, b, c, a, b, a, c, b, c, a, b, c; 8 subsets.

c. you get exactly the same list as the subsets on the right column.

2 Selected Answers to the Exercises

Chapter One Exercises

1.1 Exercises

1. These are found in the Key Concepts. 2. b. ‖u‖  65 ; c. u1  1

65 ⟨−4, 7 and u2  −1

65 ⟨−4, 7 d. 3v  ⟨9, 15,

5w  ⟨5,−10, v  5w  ⟨14, 5 and 3v− 5w  ⟨4, 25 3. b. 2u  ⟨10,−6, 4, 3w  ⟨−6, 15, 12, 2u  3w  ⟨4, 9, 16 and 2u − 3w  ⟨16,−21,−8

c. ‖w‖  45  3 5 d. u1  1 3 5

⟨−2, 5, 4 and u1  −1 3 5

⟨−2, 5, 4.

e. i. − 3 5

w  ⟨6/5,−3,−12/5 ii. 2u  5v  ⟨30,−6,−31 iii. 3w − 4u  ⟨−26, 27, 4 iv. −4u  7v− 2w  ⟨12, 2,−65.

4. a. u  v  ⟨1,−2, 7, 3 b. u  w  ⟨−1,−3, 4,−2 c. v− w  ⟨2, 1, 3, 5 d. − 2u  ⟨−6, 10,−2,−14 e. 3

4 v  − 3

2 , 9

4 , 9

2 ,−3 f. − 5

3 w  20

3 ,− 10

3 ,−5, 15

g. 5u  3v  ⟨9,−16, 23, 23 h. − 3 2

u  5 4

v  −7, 45 4

, 6,− 31 2

i. 2u − 3v  7w  ⟨−16,−5, 5,−37 j. − 5u  2v− 4w  ⟨−3, 23,−5,−7 k. − 3

2 u  3

4 v− 5

3 w  2

3 , 77

12 ,−2, 3

2 l. 3

2 u − 3

4 v  2w  −2,− 23

4 , 3,− 9

2 5. u  ⟨−15, 6, 7 and v  ⟨42,−17,−16. 6. Yes: ⟨−3, 7  40⟨5,−2  29⟨−7, 3. 7. Yes: ⟨−17,−9, 29,−37  5⟨3,−5, 1, 7  8⟨−4, 2, 3,−9. 8. No: Using the first two coordinates, we get x  −4 and y  9, but although these satisfy

the 3rd coordinate, they do not satisfy the 4th. 9. u  ⟨−3, 4, 2, 6,−7 and v  ⟨−1,−3, 5,−3, 2. 10. 7,−3 11. −4, 1, 7 12. u  ⟨−4, 4,−8 22. Contrapositive: if u  ⟨u1, u2 and v  ⟨v1, v2 are vectors in 2, then they are not

parallel to each other if and only if u1v2 − u2v1 ≠ 0. 35. PQ is 26 cm. long.

1.2 Exercises

1. y  4x/7 2. y  −5x/3 3. x  5t, y  −4t, z  2t, and t  x/5  y/−4  z/2. 4. x  −t, y  3t, z  −6t, and t  −x  y/3  z/−6. 5. 7x  5y  6 6. x  2 − 3t, y  −7  6t, z  4  8t, and t  x − 2

−3 

y  7 6

 z − 4 8

7. x  3  2t, y  2, z  −5 − 5t. Not possible because the direction vector has 0 in the y-component.

8. v  PQ  ⟨4,−2, 3, so x  −4  4t, y  3 − 2t, z  −5  3t is one possible answer (other answers are possible).

9. 2x − 11y  z  0. 10. 31x − 29y − 13z  0. 11. 10x − 2y  15z  0. We must solve for s from y, solve for r from z, then substitute these

into x. 12. Span⟨4,−10, 6,⟨−6, 15,−9 is only a line through the origin, because the vectors are

parallel to each other. 13. x  y  z  3 14. 9x  10y − 2z  28

Selected Answers to the Exercises 3

15. They determine a line because the vector AB is parallel to AC. 17. 3,−4, 7 satisfies the equation. If t  1, we get the point 7,−7, 13, which also satisfies

the equation. Since two points on the line are also on the plane, the whole line is on the plane. Alternatively, you can solve for x, y and z from the equation of the line, and substitute them into that of the plane, and get 0  0, showing that the equation of the plane is satisfied by every point on the line.

18. If Q  3, 4,−1, then PQ  ⟨1,−1,−8 is not parallel to ⟨1,−2, 5, so P is not on L. Equation: 21x  13y  z  114.

19. 7x  2y  4z  15 20. − 65 29

, 21 29

, 52 29

23. a. the point does not satisfy the symmetric equations; b. x − 5 3

 y  2

5  −z  4

24. 13x − 7y  4z  95 25. 17x − 4y  22z  −80 28. 28. a. 2x  6y  3z  0; b. w does not satisfy this equation. 33. d. 6x − 5y  4z  60; g. 3x − 2z  18 i. z  −5 34. a. D  at  x0 − x12  bt  y0 − y12  ct  z0 − z12

b. dD dt

 2t  ax0 − x1  by0 − y1  cz0 − z1

c. t  ax1 − x0  by1 − y0  cz1 − z0; d. d 2D

dt2  2  0.

e. the critical point is a local minimum by the 2nd derivative test; since D goes to positive infinity in both directions, the critical point is also an absolute maximum.

35. The critical value is t  −3 66

; 53 11

,− 67 22

, 51 22

; distance: 7 22

374

36. The critical value is t  14 30

; 98 15

, 7 3

,− 46 15

; distance: 1 15

25530

1.3 Exercises

1. ‖u‖  119 . 2. 2. cos  5/ 3161 and  ≈ 1. 481 radians. 3. ‖2u  5v‖  941 ≈ 32. 68, and ‖2u‖  ‖5v‖  136  1625 ≈ 51. 97. The second

quantity should be bigger by the Triangle Inequality. 4. cos  37/ 6391 , so  ≈ 1. 09 radians. 5. cos  −15/ 7 23 , so  ≈ 2. 034 radians. 6. cos−1 1/ 3  54. 73560

7. 71. 06820, 60. 87840, 35. 79580

8. −2911 9. 4569 10. 7837 11. 24 12. ‖u‖  29, ‖v‖  13, and ‖4u  9v‖  2305 13. 6x − 5y  2z  −15. 14. 2x  5y − 9z  40 15. Take the dot product with both u and v. 16. a. 13, 3, 6 b. 5x  13y  z  110 17. x  y − z  10; they intersect at 2, 5,−3.

4 Selected Answers to the Exercises

18. ⟨x, y, z  ⟨5,−3, 7  t⟨9, 22, 17; they intersect at 97 14

, 12 7

, 149 14

.

19. c. 7x  5y − 3z  50 20. c. 7x  11y − 13z  46 and 7x  11y − 13z  104 21. ⟨3,−5, 2 ∘ ⟨2, 4, 7  0; ⟨x, y, z  ⟨ 9

22 ,− 21

22 , 0  t⟨−43,−17, 22;

22. 4x  y − z  20 23. b. 15x  13y  10z  68; c. ⟨x, y, z  ⟨3, 1, 1  t⟨2, 0,−3 24. The direction vector of L is a multiple of the normal vector to . 25. 8x  5y − 4z  2; they intersect at 118

105 , 62

21 , 571

105 26. ⟨x, y, z  ⟨5,−2, 1  t⟨3, 7,−4; they intersect at 397

74 ,− 85

74 , 19

37 27. b. x  2z  12. 28. False: the converse is True, but the forward implication is False; u ∘ v  0 means the two

vectors are orthogonal to each other without one of them necessarily being 0 n.

1.4 Exercises

1. ⟨−3, 2, 6; all variables are leading 2. ⟨−9, 4, 0; all variables are leading 3. ⟨−3 − 7r, 2  4r, r, x3  r is free 4. ⟨6  3r, r,−7; x2  r is free 5. ⟨2,−5, r; x3  r is free 6. ⟨8  5r − 2s, r, s; x2  r and x3  s are free 7. ⟨3  5r,−4r,−2  7r, r; x4  r is free 8. ⟨5 − 3r, 6  2r, r,−4; x3  r is free 9. no solutions 10. ⟨5  4r, r,−3 − s, s; x2  r and x4  s are free 11. ⟨7  2r − 6s, r, s,−2; x2  r and x3  s are free 12.

5 3  2

3 r,− 7

3 − 4

3 r, 2

3 − 1

3 r, r ; x4  r is free

13. ⟨−5, 3, 2; all variables are leading 14. ⟨2 − 3r,−4  5r, r; x3  r is free 15. ⟨7  6r, r,−2; x2  r is free 16. ⟨5, 6,−4, 0; all variables are leading 17. ⟨4r, 3 − 7r,−8 − 3r, r; x4  r is free 18. ⟨1  6r, 5 − 4r, r,−4; x3  r is free 19. ⟨−2  5r, r, 3, 7; x2  r is free 20. ⟨−8  3r − 2s,−5 − 4r  6s, r, s; x3  r and x4  s

are free 21. ⟨−2  5r  9s, r,−6 − 4s, s; x2  r and x4  s are free 22. ⟨−5 − 7r − 5s, 2  4r − 3s, 4 − 6r  2s, r, s; x4  r and x5  s are free 23. ⟨5 − 3r  4s  6t,−1  2r  9s − 8t, r, s, t; x3  r, x4  s and x5  t are free. 24. ⟨−5 − 6r, 2  3r, 4 − 2r,−1 − 8r, r; x5  r is free 25. ⟨5 − 3r, 6  2r, r,−4, 9; x3  r is

free 26. ⟨2 − 6r − 3s, r, 7  8s, s,−3; x2  r and x4  s are free 27. ⟨−2  5r − 4s, r, 9 − 7s, 6 − 3s, s; x2  r and x5  s are free 28. no solutions 29. ⟨r, 2  3s, s,−7, 4; x1  r and x3  s are free 30. ⟨4  5r − 3s, 5 − 3r,−2  2r − 4s, 3 − 7r  6s, r, s; x5  r and x6  s are free 31. ⟨7  9r − 4s,−3r  s, r,−1 − 6s, 2 − 5s, s; x3  r and x6  s are free 32. ⟨2 − 6r − 3s − 5t, r, 9  8s  2t, s, t,−1, x2  r, x4  s and x5  t are free 33. ⟨3 − 5r,−7  2r, r, 9, 4, x3  r is free 34. ⟨−2  4r − 7s, 5 − 6r  3s, r, 6 − 9s, s,

x3  r, x5  s are free 35. ⟨−2  8r  s, 6 − 5r − 7s, r, 3  4s, 8 − 9s, s, x3  r, x6  s are free 36. ⟨−5 − 6r, 2  7r, 3 − 4r, r,−8, 9, x4  r is free 37. Yes, b  3v1 − 5v2 (only solution) 38. b is not in SpanS. 39. Yes. b  3v1 − 2v2  v3 (only solution) 40. Yes. b  1

2 v1  32 v2 (there are infinitely many solutions) 41. b

 is not in SpanS.

Selected Answers to the Exercises 5

42. Yes. b  5v1 − 2v2  4v3 (only solution) 43. Yes. b  −17v1  13v2 (there are infinitely many solutions)

44. Yes. b  3v1 − 2v2  5v3 (there are infinitely many solutions) 45. Yes. b  −2v1  5v2 (there are infinitely many solutions) 46. Yes. b  2v1 − 7v2  3v4 (there are infinitely many solutions) 47. Yes.

b  v1 − v2  2v3 (only solution) 48. Yes. b  5v1 − 4v2 (there are infinitely many solutions) 49. 0, 27

,− 3 7

50. 43 11

,− 8 11

,− 8 11

, 2 11

51. − 7 5

,− 8 5

,− 8 5

,− 7 5

52. −2s, 6s − 47 3

, 8 3

, s , where

x4  s ∈ . 53. ⟨−7,−1,−26, 31, 2, 7 54. 8 − 6r − 17t

4 , r,−7,−2 − t

4 , t,−1 , where x5  t ∈ .

55. 8 − 9s,− 1 4  25

4 s, 5 − 5s

4 , 3  4s, 8 − 9s, s , where x6  s ∈ . 56. ⟨5,−3,−9 57.

⟨11,−3, 4 58. ⟨−14,−2, 3, 2 59. ⟨−3 − 3r  4s, r,−2 − 2s, s, 2, x2  r ∈ , x4  s ∈  are free. 60. ⟨3  5r, r,−2, 4, y  r ∈  is free. 61. ⟨3 − 5r,−7  2r, r, 4, z  r ∈  is free. 62. No solutions. 63. One possible answer: ⟨x, y, z  ⟨40, 22, 0  t⟨−43,−25, 2. 64. $1.50 per shirt, $5 per pair of slacks, and $7 per jacket. 65. 1 kilogram of Barley, 3 kilograms of Oats, and 2 kilogram of Soy.

66. The rref is 1 0 − 4

5 159 5

0 1 9 5

331 5

, so d  159  4p/5 and n  331 − 9p/5.

The solution with the smallest number of pennies has p  4, n  59, and d  35. (Note: since we want n ≥ 0, we need p ≤ 36) The solution with the largest number of pennies has p  34, n  5 and d  59.

1.5 Exercises

1. a. consistent, and b. square 2. a. consistent, and b. overdetermined 3. a. inconsistent, and b. overdetermined 4. a. consistent, and b. underdetermined 5. a. inconsistent, and b. underdetermined 6. a. consistent, and b. square 7. a. consistent, and b. square 8. a. consistent, and b. underdetermined. 9. a. consistent, and b. square. 10. a. inconsistent, and b. overdetermined. 11. independent 12. independent 13. dependent 14. dependent 15. dependent 16. independent 17. dependent 18. dependent: 2v1 − v2  v3  03.

6 Selected Answers to the Exercises

19. independent 20. independent 21. dependent: 2v1 − v2  5v3  04. 22. dependent: −4v1 − 7v2  v3  04. 23. dependent: −3v1 − v2  5v3  05. 24. dependent: −2v1 − 3v2  4v3  v4  05 25. a. − 2v1 − 3v2  v3  04 b. 5v1  7v2  v4  04 c. − v2  5v3  2v4  04 26. a. − 2v1  v2  v3  05 b. − 3v1 − 2v2  v4  05 c. − 7v2 − 3v3  2v4  05 27. a. − 3v1 − 5v2  6v3  v4  04 b. − 2v1 − 3v2  5v3  v5  04

c. − v1  7v3 − 3v4  5v5  04 28. a. − 4v1 − 5v2  v3  04 b. − 3v1 − 2v2  v4  2v5  04

c. 7v1  2v3 − 5v4 − 10v5  04 29. a. 5v1  2v2  05 b. 5v1 − 6v3  2v5  05 c. v3  v4  v5  05 30. dependent: 5 vectors in 4 must be dependent. 31. One possible dependence equation is: 32u  v − 14u  5v− 4w − 2u − v  2w  0 n. 32. The system will have no solution if r  −4 and s ≠ 7

2 . The system will have exactly one

solution if r ≠ −4 and s is any real number. The system will have an infinite number of solutions if r  −4 and s  7

2 .

33. In all cases, x is a leading variable. The system will have no solution if s  −8 and t ≠ 4. The system will have exactly one solution if s ≠ −8, t is any real number, and r ≠ −6. The system will have an infinite number of solutions involving exactly one free variable in two ways. First, if s  −8, t  4, and r ≠ −6, then y is a leading variable and z is a free variable. If r  −6, then z is automatically a leading variable because of the 2nd equation, and z  − 13

10 . This will satisfy the 3rd equation if and only if 8  s − 13

10  t − 4, so

10t  13s  −144. Thus, the second way is to have r  −6 and s and t any two real numbers satisfying 10t  13s  −144. In this case, y is a free variable. The system will never have an infinite number of solutions involving exactly two free variables.

34. c  22 46. a. False. b. False. c. True. d. False e. True. f. False. g. True. h. False. i. True. j.

False.

1.6 Exercises

1. The corresponding pairs of vectors are parallel to each other. 2. If we denote by S  v1, v2 and S/  w 1, w 2, w 3, then we will get:

v1  35 w 1  15

w 2, v2  15 w 1 − 35

w 2, w 1  32 v1  12

v2, w 2  12 v1 − 32

v2,

w 3  2v1 − v2. 3. We should apply the Equality of Spans Theorem; if S  v1, v2 and S/  w 1, w 2, then

we will get: v1  13

w 1  23 w 2, v2  53

w 1  163 w 2, w 1  8v1 − v2, w 2  − 52

v1  12 v2.

4. Although both Theorems are applicable, the first Theorem will certainly be easier to apply: corresponding pairs of vectors are parallel to each other.

5. a. S consists of 6 vectors from 3, so S is certainly dependent. b. v2 and v4 are parallel

Selected Answers to the Exercises 7

to v1. c. Eliminate v2 and v4, to get: S /  v1, v3, v5, v6 . You could also eliminate v1 and v2 and keep v4, or eliminate the v1 and v4 and keep v2. d. v5  53

v1  2v3. e.

Eliminate either v1 or v3 or v5 to get a set with 3 vectors left. One possible answer is S //  v1, v3, v6 . f. The rref of the 3  3 matrix you obtained should not have any free variables.

6. a. S consists of 5 vectors from 4, so S is certainly dependent. b. v4 is parallel to v2. c. Eliminate either v2 or v4, so one possible answer is: S /  v1, v2, v3, v5  d. v3  3v1 − 2v2 and v5  2v1  v2. e. two vectors are left; one possible answer is: S //  v1, v2 . f. the two vectors (no matter which you picked) are obviously not parallel.

7. a, b and d only. 8. a, d and e only. 9. a, b, c, d and f only. 10. S /  v1, v2, v3 ; v4  3v1  2v2 − 4v3. 11. S /  v1, v2 ; v3  3v1 − 2v2 ; v4  2v1  3v2. 12. S /  v1, v3 ; v2  −5v1; v4  3v1  5v3. 13. S /  v1, v2, v3 ; v4  3v1  4v2 − 2v3 ; v5  2v1  3v2 − v3 . 14. S /  v1, v2, v4 ; v3  4v1  7v2; v5  3v1  4v2 − 2v4 . 15. S /  v1, v3, v4 ; v2  −4v1; v5  12

v1 − 32 v3  12

v4 .

16. S /  v1, v3, v5 ; v2  3v1; v4  4v1  2v3. 17. S /  v1, v2 ; v3  4v1  3v2; v4  −v1  2v2; v5  −2v1 − v2. 18. S /  v1, v2 ; v3  2v1 − 3v2. 19. S /  v1, v2, v3 . 20. S /  v1, v2 ; v3  v1  2v2; v4  −6v1  5v2. 21. S /  v1, v2, v4 ; v3  5v1  7v2. 22. S /  v1, v3 ; v2  −3v1; v4  5v1  4v3. 23. S /  v1, v2, v4 ; v3  52

v1  92 v2; v5  v1  7v2  5v4 .

24. S /  v1, v2, v3, v5; v4  5v1  4v2 − 2v3. 25. S /  v1, v2, v3 ; v4  5v1  4v2 − 2v3; v5  7v1  5v2 − 4v3. 26. S /  v1, v2 ; v3  17

v1 − 57 v2.

27. S /  v1, v2, v3 . 28. S /  v1, v2 ; v3  47

v1  297 v2; v4  17

v1 − 57 v2.

29. S /  v1, v2, v4 ; v3  5v1  8v2. 30. S /  v1, v2, v3 ; v4  2v1 − 3v2 − 4v3. 31. S /  v1, v3, v5 ; v2  16

v1; v4  76 v1 − 9v3.

32. S /  v1, v2, v4 ; v3  −6v1  5v2; v5  5v1 − 3v2. 33. S /  v1, v2, v4, v5 ; v3  5v1  8v2. 34. S /  v1, v2, v4 ; v3  7v1 − 9v2 ; v5  2v1  v2  5v4; v6  4v1 − 6v2 − 3v4. 35. S /  v1, v3, v6 ; v2  −4v1; v4  5/3v1; v5  5/3v1  2v3. 36. S /  v1, v2; v3  3v1 − 2v2; v4  −5v2; v5  2v1  v2. 37. S /  v1, v2; v3  −2/3v1  7/3v2; v4  1/3v1  1/3v2;

v5  −1/3v1  2/3v2.

8 Selected Answers to the Exercises

38. S /  v1, v2, v4 ; v3  − 32 v2; v5  12

v2  v4.

39. S /  v1, v2, v4 ; v3  2v1 − 4v2; v5  2v1 − 3v2. 40. S /  v1, v2, v4, v5 ; v3  7v1 − 4v2; v6  6v1 − 7v2  3v4 − 5v5. 41. S /  v1, v2, v4 ; v3  5v1  8v2; v5  5v1  7v2  4v4; v6  4v1  3v2  2v4. 42. S /  v1, v2; v3  −2v1; v4  −v1  v2; v5  −2v1  5v2. 43. S /  v1, v2, v4 ; v3  4v1  3v2; v5  2v1  5v2  3v4 44. S /  v1, v2, v3, v5 ; v4  3v1  4v2 − 2v3; v6  4v1  2v2 − 3v3 − 5v5. 45. S /  v1, v2, v3, v5, v6 ; v4  5v1  3v2 − 2v3. 46. a. Two non-parallel vectors are independent. b. x  2z  0. c. only e2 is in SpanS

d. Yes, because e1 ∉ SpanS. d. No, because e2 ∈ SpanS. e. Yes, because e3 ∉ SpanS.

47. b. Yes. c. No. d. No. e. Yes. 48. b.

1 0 0 0 −2 0 0 1 −1

0 1 0 0 5 2

0 0 − 37 26

27 26

0 0 1 0 1 2

0 0 − 7 26

3 26

0 0 0 1 0 0 0 8 13

17 13

0 0 0 0 0 1 0 6 13

16 13

0 0 0 0 0 0 1 7 13

− 16 13

c. independent d. dependent e. independent f. independent g. independent 53. a. False. b. True. c. False. d. True. e. False. f. False. g. False. h. False. i. False.

1.7 Exercises

1. ⟨7, 5. 2. It doesn’t contain the origin. 3. ⟨7, 3, 0,⟨0, 4, 7 is one possibility (you can also use ⟨4, 0,−3 as a second vector. 4. ⟨5, 0, 2,⟨0, 1, 0 5. It doesn’t contain the origin. 6. v1, v2, v4 ; dimW  3 7. v1, v2, v3 ; dimW  3 8. v1, v3, v6 ; dimW  3 9. v1, v2, v4 ; dimW  3 10. v1, v2, v4, v5 ; dimW  4 11. v1, v2, v4 ; dimW  3 12. v1, v3, v6 ; dimW  3 13. v1, v2; dimW  2 14. ⟨5,−3, 6, 7, ⟨3,−1, 4, 5; dimW  2 15. ⟨5,−3, 6, 7, ⟨3,−1, 4, 5, ⟨5, 1, 8,−3; dimW  3 16. ⟨5,−3, 6, 7, ⟨3,−1, 4, 5, ⟨1, 3,−1, 1; dimW  3 17. ⟨7, 5,−4, 3, 9, ⟨4, 3,−2, 1, 5; dimW  2 18. ⟨7, 5,−4, 3, 9, ⟨4, 3,−2, 1, 5, ⟨4, 3,−5, 9, 5; dimW  3 19. ⟨7, 5,−4, 3, 9, ⟨4, 3,−2, 1, 5, ⟨4, 3,−5, 4, 5; dimW  3 20. ⟨5,−3, 7,−4, 6, 3, ⟨9,−7, 8,−9, 4, 7, ⟨4,−5,−3,−6,−7, 5; dimW  3 21. ⟨7,−3, 4, 2,−5, 2, ⟨5,−2, 3, 3,−4, 1, ⟨−4, 1,−3,−8, 5, 1; dimW  3

Selected Answers to the Exercises 9

22. ⟨7,−3, 4, 2,−5, 2, ⟨5,−2, 3, 3,−4, 1, ⟨6,−4, 3,−9,−2, 5, ⟨−4, 1,−3,−8, 5, 1; dimW  4

23. ⟨7,−3, 4, 2,−5, 2, ⟨5,−2, 3, 3,−4, 1, ⟨−4, 1,−3,−2, 4,−1; dimW  3 24. ⟨7,−3, 4, 2,−5, 2, ⟨5,−2, 3, 3,−4, 1, ⟨8,−4, 3,−9,−2, 5, ⟨−4, 1,−3,−2, 4,−1;

dimW  4 25. ⟨7,−3, 4, 2,−5, 2, ⟨5,−2, 3, 3,−4, 1, ⟨5,−3, 2,−8,−1, 4, ⟨8,−4, 3,−9,−2, 5, ⟨−4, 1,−3,−2, 4

dimW  5 26. the xz-plane. ⟨1, 0, 0, ⟨0, 0, 1; dimW  2 27. the x-axis. ⟨1, 0, 0; dimW  1 28. W is not a subspace. It is not closed under addition. 29. ⟨5, 0, 1, 0, ⟨0,−1, 0, 1; dimW  2 30. ⟨0, 5, 1, 0, 0, ⟨0, 6, 0, 1, 0, ⟨−7, 0, 0, 0, 1; dimW  3 31. It does not contain the origin. 32. W is not a subspace, because it is not closed under scalar multiplication.

1.8 Exercises

1. rowspaceA: ⟨1, 0, 0, 3,⟨0, 1, 0, 2,⟨0, 0, 1,−4; colspaceA: ⟨2,−3, 4,⟨−3, 0,−5,⟨3,−1,−2; nullspaceA: ⟨−3,−2, 4, 1; nullspaceA  03 ; rankA  3  rankA; nullityA  1; nullityA  1; 3  1  4 and 3  0  3; ⟨2,−3, 3,−12  2⟨1, 0, 0, 3 − 3⟨0, 1, 0, 2  3⟨0, 0, 1,−4 ⟨−3, 0,−1,−5  −3⟨1, 0, 0, 3 − ⟨0, 0, 1,−4; ⟨4,−5,−2, 10  4⟨1, 0, 0, 3 − 5⟨0, 1, 0, 2 − 2⟨0, 0, 1,−4

2. rowspaceA: ⟨1,−5, 0, 3,⟨0, 0, 1, 5; colspaceA: ⟨−2, 4,−3,⟨3,−2, 4; nullspaceA: ⟨5, 1, 0, 0,⟨−3, 0,−5, 1; nullspaceA: ⟨−10, 1, 8; rankA  2  rankA; nullityA  2; nullityA  1; 2  2  4 and 2  1  3; ⟨−2, 10, 3, 9  −2⟨1,−5, 0, 3  3⟨0, 0, 1, 5 ⟨4,−20,−2, 2  −4⟨1,−5, 0, 3 − 2⟨0, 0, 1, 5; ⟨−3, 15, 4, 11  −3⟨1,−5, 0, 3  4⟨0, 0, 1, 5

3. rowspaceA: ⟨1, 0, 4, 0, 3,⟨0, 1, 7, 0, 4,⟨0, 0, 0, 1,−2; colspaceA: ⟨5,−2, 3,⟨−2, 3,−4,⟨−1,−3, 2; nullspaceA: ⟨−4,−7, 1, 0, 0,⟨−3,−4, 0, 2, 1; nullspaceA  03 ; rankA  3  rankA; nullityA  2; nullityA  0; 3  2  5 and 3  0  3; ⟨5,−2, 6,−1, 9  5⟨1, 0, 4, 0, 3 − 2⟨0, 1, 7, 0, 4 − ⟨0, 0, 0, 1,−2 ⟨−2, 3, 13,−3, 12  −2⟨1, 0, 4, 0, 3  3⟨0, 1, 7, 0, 4 − 3⟨0, 0, 0, 1,−2 ⟨3,−4,−16, 2,−11  3⟨1, 0, 4, 0, 3 − 4⟨0, 1, 7, 0, 4  2⟨0, 0, 0, 1,−2

4. rowspaceA: ⟨1, 3, 0, 4, 0,⟨0, 0, 1, 2, 0,⟨0, 0, 0, 0, 1; colspaceA: ⟨−1,−3, 2,⟨−2, 3,−4,⟨5,−2, 3; nullspaceA: ⟨−3, 1, 0, 0, 0,⟨−4, 0,−2, 1, 0; nullspaceA  03 ; rankA  3  rankA; nullityA  2; nullityA  0; 3  2  5 and 3  0  3; ⟨−1,−3,−2,−8, 5  −⟨1, 3, 0, 4, 0 − 2⟨0, 0, 1, 2, 0  5⟨0, 0, 0, 0, 1 ⟨−3,−9, 3,−6,−2  −3⟨1, 3, 0, 4, 0  3⟨0, 0, 1, 2, 0 − 2⟨0, 0, 0, 0, 1

10 Selected Answers to the Exercises

⟨2, 6,−4, 0, 3  2⟨1, 3, 0, 4, 0 − 4⟨0, 0, 1, 2, 0  3⟨0, 0, 0, 0, 1 5. rowspaceA: ⟨1, 0, 4,−1,−2,⟨0, 1, 3, 2,−1; colspaceA: ⟨−2, 3,−5,⟨5,−2, 3;

nullspaceA: ⟨−4,−3, 1, 0, 0,⟨1,−2, 0, 1, 0,⟨2, 1, 0, 0, 1; nullspaceA: ⟨1, 19, 11; rankA  2  rankA; nullityA  3; nullityA  1; 2  3  5 and 2  1  3; ⟨−2, 5, 7, 12,−1  −2⟨1, 0, 4,−1,−2  5⟨0, 1, 3, 2,−1; ⟨3,−2, 6,−7,−4  3⟨1, 0, 4,−1,−2 − 2⟨0, 1, 3, 2,−1 ⟨−5, 3,−11, 11, 7  −5⟨1, 0, 4,−1,−2  3⟨0, 1, 3, 2,−1

6. rowspaceA: ⟨1, 0, 0,⟨0, 1, 0,⟨0, 0, 1; colspaceA: ⟨3, 7, 1,−9,⟨−2,−4, 0,−5,⟨5,−6, 8, 2; nullspaceA  03 ; nullspaceA: ⟨−14, 243, 141, 200; rankA  3  rankA; nullityA  0; nullityA  1; 2  2  4 and 2  2  4; ⟨3,−2, 5  3⟨1, 0, 0 − 2⟨0, 1, 0  5⟨0, 0, 1; ⟨7, 4,−6  7⟨1, 0, 0  4⟨0, 1, 0 − 6⟨0, 0, 1 ⟨1, 0, 8  1⟨1, 0, 0  8⟨0, 0, 1; ⟨−9,−5, 2  −9⟨1, 0, 0 − 5⟨0, 1, 0  2⟨0, 0, 1

7. rowspaceA: ⟨1, 0, 1,−6,⟨0, 1, 2, 5; colspaceA: ⟨2, 1,−2,−2,⟨3,−2, 1,−4; nullspaceA: ⟨−1,−2, 1, 0,⟨6,−5, 0, 1; nullspaceA: ⟨3, 8, 7, 0,⟨8,−2, 0, 7; rankA  2  rankA; nullityA  2; nullityA  2; 2  2  4 and 2  2  4; ⟨2, 3, 8, 3  2⟨1, 0, 1,−6  3⟨0, 1, 2, 5; ⟨1,−2,−3,−16  ⟨1, 0, 1,−6 − 2⟨0, 1, 2, 5 ⟨−2, 1, 0, 17  −2⟨1, 0, 1,−6  ⟨0, 1, 2, 5; ⟨−2,−4,−10,−8  −2⟨1, 0, 1,−6 − 4⟨0, 1, 2, 5

8. rowspaceA: ⟨1,−3, 0, 5,⟨0, 0, 1, 4; colspaceA: ⟨−3, 7, 5, 4,⟨1,−4, 2,−3; nullspaceA: ⟨3, 1, 0, 0,⟨−5, 0,−4, 1; nullspaceA: ⟨34, 11, 5, 0,⟨−1,−1, 0, 1; rankA  2  rankA; nullityA  2; nullityA  2; 2  2  4 and 2  2  4; ⟨−3, 9, 1,−11  −3⟨1,−3, 0, 5  ⟨0, 0, 1, 4; ⟨7,−21,−4, 19  7⟨1,−3, 0, 5 − 4⟨0, 0, 1, 4 ⟨5,−15, 2, 33  5⟨1,−3, 0, 5  2⟨0, 0, 1, 4; ⟨4,−12,−3, 8  4⟨1,−3, 0, 5 − 3⟨0, 0, 1, 4

9. rowspaceA: ⟨2, 0, 5, 0, 2,⟨0, 2, 9, 0, 14,⟨0, 0, 0, 1, 5; colspaceA: ⟨0,−7, 8,−2,⟨2, 1,−2,−2,⟨−4, 3,−1, 6; nullspaceA: ⟨−5,−9, 2, 0, 0,⟨−1,−7, 0,−5, 1; nullspaceA: ⟨4,−6,−4, 5; rankA  3  rankA; nullityA  2; nullityA  1; 3  2  5 and 3  1  4; ⟨0, 2, 9,−4,−6  ⟨0, 2, 9, 0, 14 − 4⟨0, 0, 0, 1, 5; ⟨−7, 1,−13, 3, 15  − 7

2 ⟨2, 0, 5, 0, 2  1

2 ⟨0, 2, 9, 0, 14  3⟨0, 0, 0, 1, 5

⟨8,−2, 11,−1,−11  4⟨2, 0, 5, 0, 2 − ⟨0, 2, 9, 0, 14 − ⟨0, 0, 0, 1, 5 ⟨−2,−2,−14, 6, 14  −⟨2, 0, 5, 0, 2 − ⟨0, 2, 9, 0, 14  6⟨0, 0, 0, 1, 5

10. rowspaceA: ⟨1, 0, 0, 5, 7,⟨0, 1, 0, 4, 5,⟨0, 0, 1,−2,−4; colspaceA: ⟨3, 7, 1,−9,⟨−2,−4, 0, 6,⟨5, 6, 3,−9; nullspaceA: ⟨−5,−4, 2, 1, 0,⟨−7,−5, 4, 0, 1; nullspaceA: ⟨9, 6,−6, 7; rankA  3  rankA; nullityA  2; nullityA  1; 3  2  5 and 3  1  4; ⟨3,−2, 5,−3,−9  3⟨1, 0, 0, 5, 7 − 2⟨0, 1, 0, 4, 5  5⟨0, 0, 1,−2,−4 ⟨7,−4, 6, 7, 5  7⟨1, 0, 0, 5, 7 − 4⟨0, 1, 0, 4, 5  6⟨0, 0, 1,−2,−4 ⟨1, 0, 3,−1,−5  ⟨1, 0, 0, 5, 7  3⟨0, 0, 1,−2,−4; ⟨−9, 6,−9,−3, 3  −9⟨1, 0, 0, 5, 7  6⟨0, 1, 0, 4, 5 − 9⟨0, 0, 1,−2,−4

11. rowspaceA: ⟨7, 0, 4, 1,⟨0, 7, 29,−5; colspaceA: ⟨15,−3, 13,−9,−11,⟨3,−2, 4, 1, 2; nullspaceA: ⟨−4,−29, 7, 0,⟨−1, 5, 0, 7; nullspaceA: ⟨−2, 3, 3, 0, 0,⟨1, 2, 0, 1, 0,⟨4, 9, 0, 0, 3; rankA  2  rankA; nullityA  2; nullityA  3; 2  2  4 and 2  3  5;

Selected Answers to the Exercises 11

⟨15, 3, 21, 0  15 7 ⟨7, 0, 4, 1  3

7 ⟨0, 7, 29,−5;

⟨−3,−2,−10, 1  −3 7 ⟨7, 0, 4, 1 − 2

7 ⟨0, 7, 29,−5

⟨13, 4, 24,−1  13 7 ⟨7, 0, 4, 1  4

7 ⟨0, 7, 29,−5;

⟨−9, 1,−1,−2  −9 7 ⟨7, 0, 4, 1  1

7 ⟨0, 7, 29,−5

⟨−11, 2, 2,−3  −11 7

⟨7, 0, 4, 1  2 7 ⟨0, 7, 29,−5

12. rowspaceA: ⟨1, 0, 5, 0,⟨0, 1, 8, 0,⟨0, 0, 0, 1; colspaceA: ⟨3,−2,−1, 2,⟨7,−4, 3, 6,⟨1, 0, 5, 1; nullspaceA: ⟨−5,−8, 1, 0; nullspaceA: ⟨−1, 2,−2, 1, 0,⟨23,−13, 14, 0, 2; rankA  3  rankA; nullityA  1; nullityA  2; 3  1  4 and 3  2  5; ⟨3,−2,−1, 2  3⟨1, 0, 5, 0 − 2⟨0, 1, 8, 0  2⟨0, 0, 0, 1; ⟨7,−4, 3, 6  7⟨1, 0, 5, 0 − 4⟨0, 1, 8, 0  6⟨0, 0, 0, 1 ⟨1, 0, 5, 1  ⟨1, 0, 5, 0  ⟨0, 0, 0, 1; ⟨−9, 6, 3,−8  −9⟨1, 0, 5, 0  6⟨0, 1, 8, 0 − 8⟨0, 0, 0, 1 ⟨4,−3,−4, 9  4⟨1, 0, 5, 0 − 3⟨0, 1, 8, 0  9⟨0, 0, 0, 1

13. rowspaceA: ⟨1, 0, 0, 2,⟨0, 1, 0,−3,⟨0, 0, 1,−4; colspaceA: ⟨5,−3, 3,−9,−1,⟨3,−2, 4, 1, 2,⟨2,−1, 3,−1, 2; nullspaceA: ⟨−2, 3, 4, 1; nullspaceA: ⟨30, 29,−9, 4,⟨2,−3,−5, 0, 4; rankA  3  rankA; nullityA  1; nullityA  2; 3  1  4 and 3  2  5; ⟨5, 3, 2,−7  5⟨1, 0, 0, 2  3⟨0, 1, 0,−3  2⟨0, 0, 1,−4; ⟨−3,−2,−1, 4  −3⟨1, 0, 0, 2 − 2⟨0, 1, 0,−3 − ⟨0, 0, 1,−4 ⟨3, 4, 3,−18  3⟨1, 0, 0, 2  4⟨0, 1, 0,−3  3⟨0, 0, 1,−4; ⟨−9, 1,−1,−17  −9⟨1, 0, 0, 2  ⟨0, 1, 0,−3 − ⟨0, 0, 1,−4 ⟨−1, 2, 2,−16  −1⟨1, 0, 0, 2  2⟨0, 1, 0,−3  2⟨0, 0, 1,−4

14. rowspaceA: ⟨6, 1, 0, 7, 0,⟨0, 0, 1,−9, 0,⟨0, 0, 0, 0, 1; colspaceA: ⟨12,−6, 18,−6, 12,⟨3,−2, 4, 1, 2,⟨5,−3, 0, 7,−1; nullspaceA: ⟨−1, 6, 0, 0, 0,⟨−7, 0, 54, 6, 0; nullspaceA  Span⟨1, 4, 1, 1, 0,⟨−4,−9,−5, 0, 7; rankA  3  rankA; nullityA  2; nullityA  2; 3  2  5 and 3  2  5; ⟨12, 2, 3,−13, 5  2⟨6, 1, 0, 7, 0  3⟨0, 0, 1,−9, 0  5⟨0, 0, 0, 0, 1 ⟨−6,−1,−2, 11,−3  −⟨6, 1, 0, 7, 0 − 2⟨0, 0, 1,−9, 0 − 3⟨0, 0, 0, 0, 1 ⟨18, 3, 4,−15, 0  3⟨6, 1, 0, 7, 0  4⟨0, 0, 1,−9, 0; ⟨−6,−1, 1,−16, 7  −6⟨6, 1, 0, 7, 0  ⟨0, 0, 1,−9, 0  7⟨0, 0, 0, 0, 1 ⟨12, 2, 2,−4,−1  2 1, 1

6 , 0, 7

6 , 0  2⟨0, 0, 1,−9, 0 − ⟨0, 0, 0, 0, 1

15. rowspaceA: ⟨1, 0, 5, 0, 0,⟨0, 1, 8, 0, 0,⟨0, 0, 0, 1, 0,⟨0, 0, 0, 0, 1; colspaceA: ⟨3, 7, 1,−9, 4,⟨−2,−4, 0, 6,−3,⟨2, 6, 1,−8, 9,⟨5, 6, 3,−9, 7; nullspaceA: ⟨−5,−8, 1, 0, 0; nullspaceA: ⟨91, 82,−74, 93, 16; rankA  4  rankA; nullityA  1  nullityA; 4  1  5 for both matrices; ⟨3,−2,−1, 2, 5  3⟨1, 0, 5, 0, 0 − 2⟨0, 1, 8, 0, 0  2⟨0, 0, 0, 1, 0  5⟨0, 0, 0, 0, 1 ⟨7,−4, 3, 6, 6  7⟨1, 0, 5, 0, 0 − 4⟨0, 1, 8, 0, 0  6⟨0, 0, 0, 1, 0  6⟨0, 0, 0, 0, 1 ⟨1, 0, 5, 1, 3  ⟨1, 0, 5, 0, 0  5⟨0, 0, 0, 1, 0  3⟨0, 0, 0, 0, 1 ⟨−9, 6, 3,−8,−9  9⟨1, 0, 5, 0, 0 − 6⟨0, 1, 8, 0, 0 − 8⟨0, 0, 0, 1, 0 − 9⟨0, 0, 0, 0, 1 ⟨4,−3,−4, 9, 7  4⟨1, 0, 5, 0, 0 − 3⟨0, 1, 8, 0, 0  9⟨0, 0, 0, 1, 0  7⟨0, 0, 0, 0, 1

16. rowspaceA: ⟨1, 0, 7, 0, 2, 4,⟨0, 1,−9, 0, 1,−6,⟨0, 0, 0, 1, 5,−3; colspaceA: ⟨2,−1, 3,−1, 2,⟨3,−2, 4, 1, 2,⟨1,−3, 2,−2,−1;