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Solution Manual for Elementary Linear Algebra, 2nd Edition

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    Elementary Linear Algebra A Matrix Approach Spence Insel Friedberg Second Edition

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    Elementary Linear Algebra A Matrix Approach Spence Insel Friedberg Second Edition I n s t r u c t o r ’s Manual

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    Chapter 1 Matrices, Vectors, and Systems of Linear Equations 1.1 MATRICES AND VECTORS 1. » 8 − 4 20 12 16 4 – 2. » − 2 1 − 5 − 3 − 4 − 1 – 3. » 6 − 4 24 8 10 − 4 – 4. » 8 − 3 11 13 18 11 – 5. 2 4 2 4 0 6 − 4 8 3 5 6. 2 4 4 7 − 1 10 1 9 3 5 7. » 3 − 1 3 5 7 5 – 8. 2 4 4 7 − 1 10 1 9 3 5 9. 2 4 2 3 − 1 4 5 1 3 5 10. » 1 − 1 7 1 1 − 3 – 11. 2 4 − 1 − 2 0 − 3 2 − 4 3 5 12. 2 4 − 1 − 2 0 − 3 2 − 4 3 5 13. » − 3 1 − 2 − 4 − 1 − 5 6 2 – 14. 2 6 6 4 − 12 0 6 15 − 3 − 9 0 6 3 7 7 5 15. » − 6 2 − 4 − 8 − 2 − 10 12 4 – 16. » − 8 4 − 2 − 0 0 10 − 6 4 – 17. not possible 18. » 7 − 3 3 4 1 0 − 3 − 4 – 19. 2 6 6 4 7 1 − 3 0 3 − 3 4 − 4 3 7 7 5 20. » 1 1 4 12 3 25 − 24 − 2 – 21. not possible 22. 2 6 6 4 12 4 − 4 20 8 − 24 16 − 8 3 7 7 5 23. 2 6 6 4 − 7 − 1 3 0 − 3 3 − 4 4 3 7 7 5 24. 2 6 6 4 − 7 − 1 3 0 − 3 3 − 4 4 3 7 7 5 25. − 2 26. 0 27. 2 4 3 0 2 π 3 5 28. 2 4 − 2 1 . 6 5 3 5 29. » 2 2 e – 30. » 0 . 4 0 – 31. [2 − 3 0 . 4] 32. [2 e 12 0] 33. 2 4 150 150 √ 3 10 3 5 mph 34. (a) The swimmer’s velocity is u = » √ 2 √ 2 – mph.  East North . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 ◦ x y swimmer in still water Figure for Exercise 34(a) 1

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    2 Chapter 1 Matrices, Vectors, and Systems of Linear Equations (b) The water’s velocity is v = » 0 1 – mph. So the new velocity of the swimmer is u + v = » √ 2 √ 2 + 1 – mph. The correspond- ing speed is p 5 + 2 √ 2 ≈ 2 . 798 mph.             6 7 East North . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 ◦ x y water current combined velocity Figure for Exercise 34(b) 35. (a) » 150 √ 2 + 50 150 √ 2 – mph (b) 50 p 37 + 6 √ 2 ≈ 337 . 21 mph 36. The three components of the vector represent, respectively, the average blood pressure, average pulse rate, and the average cholesterol reading of the 20 people. 37. True 38. True 39. True 40. False, a scalar multiple of the zero matrix is the zero matrix. 41. False, the transpose of an m × n matrix is an n × m matrix. 42. True 43. False, the rows of B are 1 × 4 vectors. 44. False, the (3 , 4)-entry of a matrix lies in row 3 and column 4. 45. True 46. False, an m × n matrix has mn entries. 47. True 48. True 49. True 50. False, matrices must have the same size to be equal. 51. True 52. True 53. True 54. True 55. True 56. True 57. Suppose that A and B are m × n matrices. (a) The j th column of A + B and a j + b j are m × 1 vectors. The i th component of the j th column of A + B is the ( i, j )-entry of A + B , which is a ij + b ij . By definition, the i th components of a j and b j are a ij and b ij , respectively. So the i th component of a j + b j is also a ij + b ij . Thus the j th columns of A + B and a j + b j are equal. (b) The proof is similar to the proof of (a). 58. Since A is an m × n matrix, 0 A is also an m × n matrix. Because the ( i, j )-entry of 0 A is 0 a ij = 0, we see that 0 A equals the m × n zero matrix. 59. Since A is an m × n matrix, 1 A is also an m × n matrix. Because the ( i, j )-entry of 1 A is 1 a ij = a ij , we see that 1 A equals A . 60. Because both A and B are m × n matrices, both A + B and B + A are m × n matrices. The ( i, j )- entry of A + B is a ij + b ij , and the ( i, j )-entry of B + A is b ij + a ij . Since a ij + b ij = b ij + a ij by the commutative property of addition of real numbers, the ( i, j )-entries of A + B and B + A are equal for all i and j . Thus, since the matrices A + B and B + A have the same size and all pairs of corresponding entries are equal, A + B = B + A . 61. If O is the m × n zero matrix, then both A and A + O are m × n matrices; so we need only show they have equal corresponding en- tries. The ( i, j )-entry of A + O is a ij + 0 = a ij , which is the ( i, j )-entry of A . 62. The proof is similar to the proof of Exercise 61. 63. The matrices ( st ) A , tA , and s ( tA ) are all m × n matrices; so we need only show that the corre- sponding entries of ( st ) A and s ( tA ) are equal. The ( i, j )-entry of s ( tA ) is s times the ( i, j )- entry of tA , and so it equals s ( ta ij ) = st ( a ij ), which is the ( i, j )-entry of ( st ) A . Therefore ( st ) A = s ( tA ). 64. The matrices ( s + t ) A , sA , and tA are m × n ma- trices. Hence the matrices ( s + t ) A and sA + tA are m × n matrices; so we need only show they have equal corresponding entries. The ( i, j )- entry of sA + tA is the sum of the ( i, j )-entries of sA and tA , that is, sa ij + ta ij . And the ( i, j )- entry of ( s + t ) A is ( s + t ) a ij = sa ij + ta ij . 65. The matrices ( sA ) T and sA T are n × m matri- ces; so we need only show they have equal corre- sponding entries. The ( i, j )-entry of ( sA ) T is the ( j, i )-entry of sA , which is sa ji . The ( i, j )-entry of sA T is the product of s and the ( i, j )-entry of A T , which is also sa ji . 66. The matrix A T is an n × m matrix; so the ma- trix ( A T ) T is an m × n matrix. Thus we need only show that ( A T ) T and A have equal corre- sponding entries. The ( i, j )-entry of ( A T ) T is

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    1.2 Linear Combinations, Matrix-Vector Products, and Special Matrices 3 the ( j, i )-entry of A T , which in turn is the ( i, j )- entry of A . 67. If i 6 = j , then the ( i, j )-entry of a square zero matrix is 0. Because such a matrix is square, it is a diagonal matrix. 68. If B is a diagonal matrix, then B is square. Hence cB is square, and the ( i, j )-entry of cB is cb ij = c · 0 = 0 if i 6 = j . Thus cB is a diagonal matrix. 69. If B is a diagonal matrix, then B is square. Since B T is the same size as B in this case, B T is square. If i 6 = j , then the ( i, j )-entry of B T is b ji = 0. So B T is a diagonal matrix. 70. Suppose that B and C are n × n diagonal ma- trices. Then B + C is also an n × n matrix. Moreover, if i 6 = j , the ( i, j )-entry of B + C is b ij + c ij = 0 + 0 = 0. So B + C is a diagonal matrix. 71. » 2 5 5 8 – and 2 4 2 5 6 5 7 8 6 8 4 3 5 72. Let A be a symmetric matrix. Then A = A T . So the ( i, j )-entry of A equals the ( i, j )-entry of A T , which is the ( j, i )-entry of A . 73. Let O be a square zero matrix. The ( i, j )-entry of O is zero, whereas the ( i, j )-entry of O T is the ( j, i )-entry of O , which is also zero. So O = O T , and hence O is a symmetric matrix. 74. By Theorem 1.2(b), ( cB ) T = cB T = cB . 75. By Theorem 1.1(a) and Theorem 1.2(a) and (c), we have ( B + B T ) T = B T + ( B T ) T = B T + B = B + B T . 76. By Theorem 1.2(a), ( B + C ) T = B T + C T = B + C . 77. No. Consider 2 4 2 5 6 5 7 8 6 8 4 3 5 and » 2 6 5 8 – . 78. Let A be a diagonal matrix. If i 6 = j , then a ij = 0 and a ji = 0 by definition. Also, a ij = a ji if i = j . So every entry of A equals the corresponding entry of A T . Therefore A = A T . 79. The ( i, i )-entries must all equal zero. By equat- ing the ( i, i )-entries of A T and − A , we obtain a ii = − a ii , and so a ii = 0. 80. Take B = » 0 1 − 1 0 – . If C is any 2 × 2 skew- symmetric matrix, then C T = − C . Therefore c 12 = − c 21 . By Exercise 79, c 11 = c 22 = 0. So C = » 0 − c 21 c 21 0 – = − c 21 » 0 1 − 1 0 – = − c 21 B. 81. Let A 1 = 1 2 ( A + A T ) and A 2 = 1 2 ( A − A T ). It is easy to show that A = A 1 + A 2 . By Exercises 75 and 74, A 1 is symmetric. Also, by Theorem 1.2(b), (a), and (c), we have A T 2 = 1 2 ( A − A T ) T = 1 2 [ A T − ( A T ) T ] = 1 2 ( A T − A ) = − 1 2 ( A − A T ) = − A 2 . 82. (a) Because the ( i, i )-entry of A + B is a ii + b ii , we have trace( A + B ) = ( a 11 + b 11 ) + · · · + ( a nn + b nn ) = ( a 11 + · · · + a nn ) + ( b 11 + · · · + b nn ) = trace( A ) + trace( B ) . (b) The proof is similar to the proof of (a). (c) The proof is similar to the proof of (a). 83. The i th component of a p + b q is ap i + bq i , which is nonnegative. Also, the sum of the components of a p + b q is ( ap 1 + bq 1 ) + · · · + ( ap n + bq n ) = a ( p 1 + · · · + p n ) + b ( q 1 + · · · + q n ) = a (1) + b (1) = a + b = 1 . 84. (a) 2 6 6 6 6 4 6 . 5 − 0 . 5 − 1 . 9 − 2 . 8 9 . 6 − 2 . 9 1 . 5 − 3 . 0 17 . 4 0 . 4 − 15 . 5 5 . 2 − 1 . 0 − 3 . 7 − 7 . 3 17 . 5 5 . 2 1 . 4 3 . 5 16 . 8 3 7 7 7 7 5 (b) 2 6 6 6 6 4 − 1 . 3 3 . 4 − 4 . 0 10 . 4 3 . 0 4 . 9 − 2 . 4 6 . 6 − 3 . 9 − 4 . 1 9 . 4 − 8 . 6 1 . 7 − 0 . 1 − 14 . 5 − 0 . 2 − 4 . 7 4 . 1 − 0 . 7 − 1 . 8 3 7 7 7 7 5 (c) 2 6 6 4 3 . 9 7 . 4 10 . 3 − 0 . 1 1 . 9 0 . 8 − 0 . 3 − 1 . 1 − 2 . 5 2 . 3 − 2 . 6 0 . 2 − 7 . 2 − 9 . 7 2 . 1 1 . 6 0 . 2 0 . 6 11 . 6 10 . 6 3 7 7 5 1.2 LINEAR COMBINATIONS, MATRIX-VECTOR PRODUCTS, AND SPECIAL MATRICES 1. » 12 14 – 2. 2 4 − 5 4 7 3 5 3. 2 4 9 0 10 3 5 4. » 22 32 – 5. » a b – 6. [18] 7. » 22 5 – 8. 2 4 a b c 3 5

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    4 Chapter 1 Matrices, Vectors, and Systems of Linear Equations 9. 2 4 sa tb uc 3 5 10. [6] 11. 2 4 2 − 6 10 3 5 12. 2 4 − 3 4 2 3 5 13. » − 1 6 – 14. 2 4 3 − 1 2 3 5 15. » 21 13 – 16. » 26 9 – 17. 1 2 » √ 2 − √ 2 √ 2 √ 2 – , 1 2 » − √ 2 √ 2 – 18. » 1 0 0 1 – , e 1 19. 1 2 » 1 − √ 3 √ 3 1 – , 1 2 » 3 − √ 3 3 √ 3 + 1 – 20. 1 2 » √ 3 − 1 1 √ 3 – , 1 2 » √ 3 − 2 1 + 2 √ 3 – 21. 1 2 » − √ 3 1 − 1 − √ 3 – , 1 2 » √ 3 − 3 3 √ 3 + 1 – 22. − 1 √ 2 » 1 1 − 1 1 – , − 1 √ 2 » 1 − 3 – 23. » 3 2 – 24. 1 2 » 4 √ 3 + 1 √ 3 − 4 – 25. 1 2 » 3 − √ 3 3 √ 3 + 1 – 26. 1 2 » 2 − 5 √ 3 2 √ 3 + 5 – 27. 1 2 » 3 − 3 √ 3 – 28. » √ 3 1 – 29. » 1 1 – = (1) » 1 0 – + (1) » 0 1 – 30. » 1 − 1 – = 1 4 » 4 − 4 – 31. not possible 32. » 1 1 – = (1) » 1 0 – + (1) » 0 1 – 33. not possible 34. » 1 1 – = (1) » 1 0 – + ( − 1) » 0 − 1 – + 0 » 0 0 – 35. » − 1 11 – = 3 » 1 3 – + ( − 2) » 2 − 1 – 36. » 1 1 – = 0 » 1 0 – + 0 » 0 − 1 – + (1) » 1 1 – 37. » 3 8 – = 7 » 1 2 – + ( − 2) » 2 3 – + 0 » − 2 − 5 – 38. » a b – = „ a + 2 b 3 « » 1 1 – + „ a − b 3 « » 2 − 1 – 39. not possible 40. u = 4 2 4 0 1 2 3 5 + ( − 2) 2 4 − 1 3 0 3 5 41. u = 0 2 4 2 − 1 2 3 5 + 1 2 4 3 − 2 1 3 5 + 0 2 4 − 4 1 3 3 5 42. u = 5 2 4 1 0 0 3 5 + 6 2 4 0 1 0 3 5 + 7 2 4 0 0 1 3 5 43. u = ( − 4) 2 4 1 0 0 3 5 + ( − 5) 2 4 0 1 0 3 5 + ( − 6) 2 4 0 0 1 3 5 44. u = 0 2 4 1 − 1 1 3 5 + 0 2 4 0 − 2 3 3 5 + 1 2 4 − 1 3 2 3 5 45. True 46. False. If the coefficients of the linear combina- tion 3 » 2 2 – + ( − 6) » 1 1 – = » 0 0 – were positive, the sum could not equal the zero vector. 47. True 48. True 49. True 50. False, the matrix-vector product of a 2 × 3 ma- trix and a 3 × 1 vector is a 2 × 1 vector. 51. False, the matrix-vector product is a linear com- bination of the columns of the matrix. 52. False, the product of a matrix and a standard vector is a column of the matrix. 53. True 54. False, the matrix-vector product of an m × n matrix and a vector in R n yields a vector in R m . 55. False, every vector in R 2 is a linear combination of two nonparallel vectors. 56. True 57. False, a standard vector is a vector with a single component equal to 1 and the others equal to 0. 58. True 59. False, consider A = » 1 − 1 − 1 1 – and u = » 1 1 – . 60. True 61. False, A θ u is the vector obtained by rotating u by a counterclockwise rotation of the angle θ . 62. False, consider A = » 1 − 1 − 1 1 – , u = » 1 1 – , and v = » 2 2 – . 63. True 64. True 65. If θ = 0, then A θ = I 2 . So A θ v = I 2 v = v by Theorem 1.3(h). 66. We have A 180 ◦ v = » − 1 0 0 − 1 – v = − I 2 v = − v .

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    1.2 Linear Combinations, Matrix-Vector Products, and Special Matrices 5 67. Let v = » a b – . Then A θ ( A β v ) = » cos θ − sin θ sin θ cos θ – „» cos β − sin β sin β cos β – » a b –« = » cos θ − sin θ sin θ cos θ – » a cos β − b sin β a sin β + b cos β – = » a cos θ cos β − b cos θ sin β a sin θ cos β − b sin θ sin β – + » − a sin θ sin β − b sin θ cos β a cos θ sin β + b cos θ cos β – = » a cos ( θ + β ) − b sin ( θ + β ) a sin ( θ + β ) + b cos ( θ + β ) – = A θ + β v . 68. Let u = » a b – . Then A T θ ( A θ u ) = » cos θ sin θ − sin θ cos θ – „» cos θ − sin θ sin θ cos θ – » a b –« = » cos θ sin θ − sin θ cos θ – » a cos θ − b sin θ a sin θ + b cos θ – = » a cos 2 θ − b sin θ cos θ − a sin θ cos θ + b sin 2 θ – + » a sin 2 θ + b sin θ cos θ a sin θ cos θ + b cos 2 θ – = » a (sin 2 θ + cos 2 θ ) b (sin 2 θ + cos 2 θ ) – = » a b – = u . Similarly, A θ ( A T θ u ) = u . 69. (a) As in Example 3, the populations are given by the entries of A » 400 300 – = » 349 351 – ; so there will be 349,000 people in the city and 351,000 in the suburbs. (b) Computing A » 349 351 – = » 307 . 180 392 . 820 – , we see that there will be 307,180 people in the city and 392,820 in the suburbs. 70. A u = a 2 4 1 4 7 3 5 + b 2 4 2 5 8 3 5 + c 2 4 3 6 9 3 5 71. A u = » − 1 0 0 1 – » a b – = » − a b – , the reflection of u about the y -axis 72. We have A ( A u ) = A „» − 1 0 0 1 – » a b –« = » − 1 0 0 1 – » − a b – = » a b – = u . 73. B = » 1 0 0 − 1 – 74. (a) C = A 180 ◦ = » − 1 0 0 − 1 – (b) We have A ( C u ) = » − 1 0 0 1 – „» − 1 0 0 − 1 – » a b –« = » − 1 0 0 1 – » − a − b – = » a − b – . In a similar fashion, we have C ( A u ) = » a − b – = B u and B ( C u ) = C ( B u ) = A u . (c) The first equation shows that reflecting about the x -axis can be accomplished by either first rotating by 180 ◦ and then re- flecting about the y -axis, or first reflect- ing about the y -axis and then rotating by 180 ◦ . The second equation shows that reflecting about the y -axis may be accomplished ei- ther by first rotating by 180 ◦ and then re- flecting about the x -axis, or first reflect- ing about the x -axis and then rotating by 180 ◦ . 75. A u = » a 0 – , the projection of u on the x -axis 76. This exercise is similar to Exercise 72. 77. If v = » a 0 – , then A v = » 1 0 0 0 – » a 0 – = » a 0 – = v . 78. B = » 0 0 0 1 – 79. (a) We have A ( C u ) = » 1 0 0 0 – „» − 1 0 0 − 1 – » a b –« = » 1 0 0 0 – » − a − b – = » − a 0 – , and C ( A u ) = » − 1 0 0 − 1 – „» 1 0 0 0 – » a b –«

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    6 Chapter 1 Matrices, Vectors, and Systems of Linear Equations = » − 1 0 0 − 1 – » a 0 – = » − a 0 – . (b) Rotating a vector by 180 ◦ and then pro- jecting the result on the x -axis is equiva- lent to projecting a vector on the x -axis and then rotating the result by 180 ◦ . 80. The sum of the two linear combinations a u 1 + b u 2 and c u 1 + d u 2 is ( a u 1 + b u 2 )+( c u 1 + d u 2 ) = ( a + c ) u 1 +( b + d ) u 2 , which is also a linear combination of u 1 and u 2 . 81. Write v = a 1 u 1 + a 2 u 2 and w = b 1 u 1 + b 2 u 2 , where a 1 , a 2 , b 1 , and b 2 are scalars. A linear combination of v and w has the form c v + d w = c ( a 1 u 1 + a 2 u 2 ) + d ( b 1 u 1 + b 2 u 2 ) = ( ca 1 + db 1 ) u 1 + ( ca 2 + db 2 ) u 2 , which is also a linear combination of u 1 and u 2 . 82. The proof is similar to that of Exercise 81. 83. We have A ( c u ) = ( cu 1 ) a 1 + ( cu 2 ) a 2 + · · · + ( cu n ) a n = c ( u 1 a 1 + u 2 a 2 + · · · + u n a n ) = c ( A u ) . Similarly, ( cA ) u = c ( A u ). 84. We have ( A + B ) u = u 1 ( a 1 + b 1 ) + · · · + u n ( a n + b n ) = u 1 a 1 + u 1 b 1 + · · · + u n a n + u n b n = ( u 1 a 1 + · · · + u n a n ) + ( u 1 b 1 + · · · + u n b n ) = A u + B u . 85. We have A e j = 0 a 1 + · · · + 0 a j − 1 + 1 a j + 0 a j +1 + · · · + 0 a n = a j . 86. Suppose B w = A w for all w . Let w = e j . Then B e j = A e j . From Theorem 1.3(e), it follows that b j = a j for all j . So B = A . 87. The vector A 0 is an m × 1 vector. By definition A 0 = 0 a 1 + 0 a 2 + · · · + 0 a n = 0 . 88. Every column of O is the m × 1 zero vector. So O v = v 1 0 + v 2 0 + · · · + v n 0 = 0 . 89. The j th column of I n is e j . So I n v = v 1 e 1 + v 2 e 2 + · · · + v n e n = v . 90. Using p = » 400 300 – , we compute A p , A ( A p ) , . . . until we have ten vectors. From the final vector, we see that there will be 155,610 people living in the city and 544,389 people living in the suburbs after ten years. 91. (a) 2 6 6 4 24 . 6 45 . 0 26 . 0 − 41 . 4 3 7 7 5 (b) 2 6 6 4 134 . 1 44 . 4 7 . 6 104 . 8 3 7 7 5 (c) 2 6 6 4 128 . 4 80 . 6 63 . 5 25 . 8 3 7 7 5 (d) 2 6 6 4 653 . 09 399 . 77 528 . 23 − 394 . 52 3 7 7 5 1.3 SYSTEMS OF LINEAR EQUATIONS 1. (a) » 0 − 1 2 1 3 0 – (b) » 0 − 1 2 0 1 3 0 − 1 – 2. (a) ˆ 2 − 1 3 ˜ (b) ˆ 2 − 1 3 4 ˜ 3. (a) 2 4 1 2 − 1 3 − 3 4 3 5 (b) 2 4 1 2 3 − 1 3 2 − 3 4 1 3 5 4. (a) » 1 0 2 − 1 2 − 1 0 1 – (b) » 1 0 2 − 1 3 2 − 1 0 1 0 – 5. (a) 2 4 0 2 − 3 − 1 1 2 2 0 1 3 5 (b) 2 4 0 2 − 3 4 − 1 1 2 − 6 2 0 1 0 3 5 6. (a) 2 4 1 − 2 1 7 1 − 2 0 10 2 − 4 4 8 3 5 (b) 2 4 1 − 2 1 7 5 1 − 2 0 10 3 2 − 4 4 8 7 3 5 7. 2 4 0 2 − 4 4 2 − 2 6 3 − 1 1 1 − 1 0 2 − 3 3 5 8. 2 4 − 3 3 0 − 6 9 − 2 6 3 − 1 1 0 2 − 4 4 2 3 5

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    1.3 Systems of Linear Equations 7 9. 2 4 1 − 1 0 2 − 3 0 4 3 3 − 5 0 2 − 4 4 2 3 5 10. 2 4 − 2 6 3 − 1 1 1 − 1 0 2 − 3 0 2 − 4 4 2 3 5 11. 2 4 1 − 1 0 2 − 3 − 2 6 3 − 1 1 0 1 − 2 2 1 3 5 12. 2 4 1 − 1 0 2 − 3 − 2 0 15 − 13 − 5 0 2 − 4 4 2 3 5 13. 2 4 1 − 1 0 2 − 3 − 2 6 3 − 1 1 − 8 26 8 0 6 3 5 14. 2 4 1 − 1 0 2 − 3 − 2 6 3 − 1 1 2 0 − 4 8 − 4 3 5 15. 2 6 6 4 − 2 4 0 − 1 1 − 1 2 − 4 6 − 3 2 1 3 7 7 5 16. 2 6 6 6 4 1 − 2 0 − 1 2 1 2 − 1 2 2 − 4 6 − 3 2 1 3 7 7 7 5 17. 2 6 6 4 1 − 2 0 − 1 1 − 1 0 0 6 − 3 2 1 3 7 7 5 18. 2 6 6 4 1 − 2 0 − 1 1 − 1 2 − 4 6 0 − 4 1 3 7 7 5 19. 2 6 6 4 1 − 2 0 2 − 4 6 − 1 1 − 1 − 3 2 1 3 7 7 5 20. 2 6 6 4 1 − 2 0 − 3 2 1 2 − 4 6 − 1 1 − 1 3 7 7 5 21. 2 6 6 4 1 − 2 0 − 1 1 − 1 2 − 4 6 − 1 0 3 3 7 7 5 22. 2 6 6 4 − 1 0 − 2 − 1 1 − 1 2 − 4 6 − 3 2 1 3 7 7 5 23. Yes, because 1(1) − 4( − 2) + 3( − 1) = 6 and 1( − 5) − 2( − 1) = − 3. Alternatively, » 1 − 4 0 3 0 0 1 − 2 – 2 6 6 4 1 − 2 − 5 − 1 3 7 7 5 = » 6 − 3 – . 24. No, because 1(2) − 4(0) + 3(1) = 5 6 = 6. Alter- natively, if A is the coefficient matrix, and the given vector is v , then A v = » 5 − 3 – 6 = » 6 − 3 – . 25. No, because the left side of the second equation yields 1(2) − 2(1) = 0 6 = − 3. Alternatively, » 1 − 4 0 3 0 0 1 − 2 – 2 6 6 4 3 0 2 1 3 7 7 5 = » 6 0 – 6 = » 6 − 3 – . 26. Yes, the components of the vector satisfy both equations. Alternatively, if the given vector is v , then A v = » 6 − 3 – . 27. no 28. yes 29. yes 30. yes 31. yes 32. no 33. yes 34. yes 35. no 36. yes 37. no 38. no 39. x 1 = 2 + x 2 x 2 free 40. x 1 = − 4 x 2 = 5 41. x 1 = 6 + 2 x 2 x 2 free 42. x 1 = 5 + 4 x 2 x 2 free 43. not consistent 44. x 1 = − 6 x 2 = 3 45. x 1 = 4 + 2 x 2 x 2 free x 3 = 3 46. not consistent 47. x 1 = 3 x 4 x 2 = 4 x 4 x 3 = − 5 x 4 x 4 free and 2 6 6 4 x 1 x 2 x 3 x 4 3 7 7 5 = x 4 2 6 6 4 3 4 − 5 1 3 7 7 5 48. x 1 = 9 + x 3 − 3 x 4 x 2 = 8 − 2 x 3 + 5 x 4 x 3 free x 4 free and 2 6 6 4 x 1 x 2 x 3 x 4 3 7 7 5 = x 3 2 6 6 4 1 − 2 1 0 3 7 7 5 + x 4 2 6 6 4 − 3 5 0 1 3 7 7 5 + 2 6 6 4 9 8 0 0 3 7 7 5 49. x 1 free x 2 = − 3 x 3 = − 4 x 4 = 5 and 2 6 6 4 x 1 x 2 x 3 x 4 3 7 7 5 = x 1 2 6 6 4 1 0 0 0 3 7 7 5 + 2 6 6 4 0 − 3 − 4 5 3 7 7 5 50. x 1 = − 3 + 2 x 2 x 2 free x 3 = − 4 x 4 = 5 and 2 6 6 4 x 1 x 2 x 3 x 4 3 7 7 5 = x 2 2 6 6 4 2 1 0 0 3 7 7 5 + 2 6 6 4 − 3 0 − 4 5 3 7 7 5 51. x 1 = 6 − 3 x 2 + 2 x 4 x 2 free x 3 = 7 − 4 x 4 x 4 free and 2 6 6 4 x 1 x 2 x 3 x 4 3 7 7 5 = x 2 2 6 6 4 − 3 1 0 0 3 7 7 5 + x 4 2 6 6 4 2 0 − 4 1 3 7 7 5 + 2 6 6 4 6 0 7 0 3 7 7 5

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