Elementary Linear Algebra (Classic Version, 2nd Edition, William K. Spence) – Complete Solutions Manual with Detailed Problem Solving

SOLUTIONS for
Elementary Linear Algebra (Classic
Version), 2nd edition
Author (s): Lawrence E. Spence, Arnold J. Insel
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,Chapter 1

Matrices, Vectors, and
Systems of Linear
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Equations
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2 3 2 3
1.1 MATRICES AND VECTORS 12 4 −7 −1
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6−4 207 6 3 07
22. 6
4 8 −245
7 23. 6
4−3
7
»
8 −4 20
–
−2
»
1 −5
– 35
1. 2. 16 −8 −4 4
12 16 4 −3 −4 −1
2 3
»
6 −4 24
– »
8 −3 11
– −7 −1
3. 4. 6 3 07
8 10 −4 13 18 11 24. 6 7 25. −2 26. 0
2 3 2 3 4−3 35
2 4 4 7 −4 4
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5. 4 0 65 6. 4−1 105 2 3 2 3
−4 8 1 9 3 −2 » –
405 2
27. 28. 41.65 29.
2
4 7
3 2e
»
3 −1 3
– 2π 5
7. 8. 4−1 105 » –
5 7 5 0.4
1 9 30. 31. [2 − 3 0.4] 32. [2e 12 0]
0
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2 3
2 3 » –
−1
2 3
1 7 150
9. 4−1 45 10. √
1 1 −3 33. 4150 35 mph
5 1
2 3 2 3 10
−1 −2 −1 −2 »√ –
−35 −35 2
11. 4 0 12. 4 0 34. (a) The swimmer’s velocity is u = √ mph.
2 −4 2 −4 2
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2 3
» – −12 0
−3 1 −2 −4 6 6 157
13. 14. 6 7
−1 −5 6 2 4 −3 −95 North
0 6 y
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» –
−6 2 −4 −8
15. 
−2 −10 12 4 swimmer in
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» – still water
−8 4 −2 −0
16. 17. not possible
0 10 −6 4 .... ◦
...45
2
7 1
3 ...
.
x
» – East
7 −3 3 4 6−3 07
18. 19. 4
6 7
1 0 −3 −4 3 −35 Figure for Exercise 34(a)
4 −4
» –
1 1 4 12
20. 21. not possible
3 25 −24 −2


1

,2 Chapter 1 Matrices, Vectors, and Systems of Linear Equations

» –
0 A+B, which is aij +bij . By definition, the
(b) The water’s velocity is v = mph. So
1 ith components of aj and bj are aij and
the new velocity
» √ of – the swimmer is bij , respectively. So the ith component of
2 aj + bj is also aij + bij . Thus the jth
u+v = √ mph. The correspond-
2+1 columns of A + B and aj + bj are equal.
p √
ing speed is 5 + 2 2 ≈ 2.798 mph. (b) The proof is similar to the proof of (a).
58. Since A is an m × n matrix, 0A is also an m × n
North
y matrix. Because the (i, j)-entry of 0A is 0aij =
combined 7 6 0, we see that 0A equals the m × n zero matrix.
velocity 
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59. Since A is an m × n matrix, 1A is also an m × n
 water current
 matrix. Because the (i, j)-entry of 1A is 1aij =
 aij , we see that 1A 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 aij + bij , and the (i, j)-entry
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.
.... ◦ of B + A is bij + aij . Since aij + bij = bij + aij
 ..45
 .... x by the commutative property of addition of real
East
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numbers, the (i, j)-entries of A+B and B+A are
equal for all i and j. Thus, since the matrices
Figure for Exercise 34(b)
A + B and B + A have the same size and all
» √ – pairs of corresponding entries are equal, A+B =
150 2√+ 50
35. (a) mph B + A.
150 2
p √ 61. If O is the m × n zero matrix, then both A
(b) 50 37 + 6 2 ≈ 337.21 mph and A + O are m × n matrices; so we need
36. The three components of the vector represent, only show they have equal corresponding en-
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respectively, the average blood pressure, average tries. The (i, j)-entry of A + O is aij + 0 = aij ,
pulse rate, and the average cholesterol reading which is the (i, j)-entry of A.
of the 20 people. 62. The proof is similar to the proof of Exercise 61.
37. True 38. True 39. True 63. The matrices (st)A, tA, and s(tA) are all m × n
matrices; so we need only show that the corre-
40. False, a scalar multiple of the zero matrix is the
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sponding entries of (st)A and s(tA) are equal.
zero matrix.
The (i, j)-entry of s(tA) is s times the (i, j)-
41. False, the transpose of an m × n matrix is an entry of tA, and so it equals s(taij ) = st(aij ),
n × m matrix. which is the (i, j)-entry of (st)A. Therefore
42. True (st)A = s(tA).
43. False, the rows of B are 1 × 4 vectors. 64. The matrices (s+t)A, sA, and tA are m×n ma-
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trices. Hence the matrices (s + t)A and sA + tA
44. False, the (3, 4)-entry of a matrix lies in row 3 are m × n matrices; so we need only show they
and column 4. have equal corresponding entries. The (i, j)-
45. True entry of sA + tA is the sum of the (i, j)-entries
of sA and tA, that is, saij + taij . And the (i, j)-
46. False, an m × n matrix has mn entries.
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entry of (s + t)A is (s + t)aij = saij + taij .
47. True 48. True 49. True
65. The matrices (sA)T and sAT are n × m matri-
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50. False, matrices must have the same size to be ces; so we need only show they have equal corre-
equal. sponding entries. The (i, j)-entry of (sA)T is the
51. True 52. True 53. True (j, i)-entry of sA, which is saji . The (i, j)-entry
of sAT is the product of s and the (i, j)-entry of
54. True 55. True 56. True AT , which is also saji .
57. Suppose that A and B are m × n matrices. 66. The matrix AT is an n × m matrix; so the ma-
(a) The jth column of A + B and aj + bj are trix (AT )T is an m × n matrix. Thus we need
m × 1 vectors. The ith component of the only show that (AT )T and A have equal corre-
jth column of A + B is the (i, j)-entry of sponding entries. The (i, j)-entry of (AT )T is

, 1.2 Linear Combinations, Matrix-Vector Products, and Special Matrices 3


the (j, i)-entry of AT , which in turn is the (i, j)- 81. Let A1 = 12 (A + AT ) and A2 = 21 (A − AT ). It
entry of A. is easy to show that A = A1 + A2 . By Exercises
67. If i 6= j, then the (i, j)-entry of a square zero 75 and 74, A1 is symmetric. Also, by Theorem
matrix is 0. Because such a matrix is square, it 1.2(b), (a), and (c), we have
is a diagonal matrix. 1 1
68. If B is a diagonal matrix, then B is square. AT2 = (A − AT )T = [AT − (AT )T ]
2 2
Hence cB is square, and the (i, j)-entry of cB 1 T 1
is cbij = c · 0 = 0 if i 6= j. Thus cB is a diagonal = (A − A) = − (A − AT ) = −A2 .
2 2
matrix.
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69. If B is a diagonal matrix, then B is square. Since 82. (a) Because the (i, i)-entry of A+B is aii +bii ,
B T is the same size as B in this case, B T is we have
square. If i 6= j, then the (i, j)-entry of B T is trace(A + B)
bji = 0. So B T is a diagonal matrix. = (a11 + b11 ) + · · · + (ann + bnn )
70. Suppose that B and C are n × n diagonal ma- = (a11 + · · · + ann ) + (b11 + · · · + bnn )
trices. Then B + C is also an n × n matrix.
= trace(A) + trace(B).
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Moreover, if i 6= j, the (i, j)-entry of B + C is
bij + cij = 0 + 0 = 0. So B + C is a diagonal (b) The proof is similar to the proof of (a).
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matrix. 2 3 (c) The proof is similar to the proof of (a).
» – 2 5 6
2 5 83. The ith component of ap + bq is api + bqi , which
71. and 45 7 85
5 8 is nonnegative. Also, the sum of the components
6 8 4
of ap + bq is
72. Let A be a symmetric matrix. Then A = AT .
So the (i, j)-entry of A equals the (i, j)-entry of (ap1 + bq1 ) + · · · + (apn + bqn )
AT , which is the (j, i)-entry of A. = a(p1 + · · · + pn ) + b(q1 + · · · + qn )
73. Let O be a square zero matrix. The (i, j)-entry
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= a(1) + b(1) = a + b = 1.
of O is zero, whereas the (i, j)-entry of OT is the
(j, i)-entry of O, which is also zero. So O = OT ,
2 3
6.5 −0.5 −1.9 −2.8
and hence O is a symmetric matrix. 6 9.6 −2.9 1.5 −3.07
6 7
74. By Theorem 1.2(b), (cB)T = cB T = cB. 84. 6 17.4
(a) 6 0.4 −15.5 5.27
7
75. By Theorem 1.1(a) and Theorem 1.2(a) and (c), 4−1.0 −3.7 −7.3 17.55
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we have 5.2 1.4 3.5 16.8
2 3
(B + B T )T = B T + (B T )T = B T + B = B + B T . −1.3 3.4 −4.0 10.4
6 3.0 4.9 −2.4 6.67
76. By Theorem 1.2(a), (B + C)T = B T + C T =
6 7
6−3.9
(b) 6 −4.1 9.4 −8.67
7
B + C. 4 1.7 −0.1 −14.5 −0.25
2 3
2 5 6 » – −4.7 4.1 −0.7 −1.8
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2 6
77. No. Consider 5 7 8 and
4 5 . 2 3
5 8 3.9 7.4 10.3 −0.1 1.9
6 8 4 6 0.8 −0.3 −1.1 −2.5 2.37
78. Let A be a diagonal matrix. If i 6= j, then aij = (c) 4
6 7
−2.6 0.2 −7.2 −9.7 2.15
0 and aji = 0 by definition. Also, aij = aji if i = 1.6 0.2 0.6 11.6 10.6
j. So every entry of A equals the corresponding
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entry of AT . Therefore A = AT .
1.2 LINEAR COMBINATIONS,
79. The (i, i)-entries must all equal zero. By equat-
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ing the (i, i)-entries of AT and −A, we obtain MATRIX-VECTOR PRODUCTS,
aii = −aii , and so aii = 0. AND SPECIAL MATRICES
» –
0 1 2 3 23
80. Take B = . If C is any 2 × 2 skew- » – −5 9 » –
−1 0 12 22
1. 2. 4 45 3. 4 05 4.
symmetric matrix, then C T = −C. Therefore 14
7 10
32
c12 = −c21 . By Exercise 79, c11 = c22 = 0. So 2 3
» – » – » – »– a
0 −c21 0 1 a 22
C= = −c21 = −c21 B. 5. 6. [18] 7. 8. 4 b 5
c21 0 −1 0 b 5
c