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Solution Manual for Elementary Linear Algebra with Applications (Classic Version), 9th Edition

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    Solutions Manual Elementary Linear Algebra with Applications Ninth Edition Bernard Kolman Drexel University David R. Hill Temple University

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    Contents Preface iii 1 Linear Equations and Matrices 1 1.1 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.4 Algebraic Properties of Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5 Special Types of Matrices and Partitioned Matrices . . . . . . . . . . . . . . . . . . . . . . . . 9 1.6 Matrix Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.7 Computer Graphics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.8 Correlation Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2 Solving Linear Systems 27 2.1 Echelon Form of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.2 Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.3 Elementary Matrices; Finding A − 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2.4 Equivalent Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.5 LU -Factorization (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3 Determinants 37 3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.2 Properties of Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.3 Cofactor Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.4 Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.5 Other Applications of Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4 Real Vector Spaces 45 4.1 Vectors in the Plane and in 3-Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 4.2 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.3 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.4 Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 4.5 Span and Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 4.6 Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 4.7 Homogeneous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 4.8 Coordinates and Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 4.9 Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

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    ii CONTENTS Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 5 Inner Product Spaces 71 5.1 Standard Inner Product on R 2 and R 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 5.2 Cross Product in R 3 (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 5.3 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 5.4 Gram-Schmidt Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 5.5 Orthogonal Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 5.6 Least Squares (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 6 Linear Transformations and Matrices 93 6.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 6.2 Kernel and Range of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 6.3 Matrix of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 6.4 Vector Space of Matrices and Vector Space of Linear Transformations (Optional) . . . . . . . 99 6.5 Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 6.6 Introduction to Homogeneous Coordinates (Optional) . . . . . . . . . . . . . . . . . . . . . . 103 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 7 Eigenvalues and Eigenvectors 109 7.1 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 7.2 Diagonalization and Similar Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 7.3 Diagonalization of Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 8 Applications of Eigenvalues and Eigenvectors (Optional) 129 8.1 Stable Age Distribution in a Population; Markov Processes . . . . . . . . . . . . . . . . . . . 129 8.2 Spectral Decomposition and Singular Value Decomposition . . . . . . . . . . . . . . . . . . . 130 8.3 Dominant Eigenvalue and Principal Component Analysis . . . . . . . . . . . . . . . . . . . . 130 8.4 Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 8.5 Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 8.6 Real Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 8.7 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 8.8 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 10 MATLAB Exercises 137 Appendix B Complex Numbers 163 B.1 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 B.2 Complex Numbers in Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

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    Chapter 1 Linear Equations and Matrices Section 1.1, p. 8 2. x = 1, y = 2, z = − 2. 4. No solution. 6. x = 13 + 10 t , y = − 8 − 8 t , t any real number. 8. Inconsistent; no solution. 10. x = 2, y = − 1. 12. No solution. 14. x = − 1, y = 2, z = − 2. 16. (a) For example: s = 0, t = 0 is one answer. (b) For example: s = 3, t = 4 is one answer. (c) s = t 2 . 18. Yes. The trivial solution is always a solution to a homogeneous system. 20. x = 1, y = 1, z = 4. 22. r = − 3. 24. If x 1 = s 1 , x 2 = s 2 , . . . , x n = s n satisfy each equation of (2) in the original order, then those same numbers satisfy each equation of (2) when the equations are listed with one of the original ones interchanged, and conversely. 25. If x 1 = s 1 , x 2 = s 2 , . . . , x n = s n is a solution to (2), then the p th and q th equations are satisfied. That is, a p 1 s 1 + · · · + a pn s n = b p a q 1 s 1 + · · · + a qn s n = b q . Thus, for any real number r , ( a p 1 + ra q 1 ) s 1 + · · · + ( a pn + ra qn ) s n = b p + rb q . Then if the q th equation in (2) is replaced by the preceding equation, the values x 1 = s 1 , x 2 = s 2 , . . . , x n = s n are a solution to the new linear system since they satisfy each of the equations.

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    2 Chapter 1 26. (a) A unique point. (b) There are infinitely many points. (c) No points simultaneously lie in all three planes. 28. No points of intersection: C 1 C 2 C 2 C 1 One point of intersection: C 1 C 2 Two points of intersection: C 1 C 2 Infinitely many points of intersection: C 1 C 2 = 30. 20 tons of low-sulfur fuel, 20 tons of high-sulfur fuel. 32. 3.2 ounces of food A, 4.2 ounces of food B, and 2 ounces of food C. 34. (a) p (1) = a (1) 2 + b (1) + c = a + b + c = − 5 p ( − 1) = a ( − 1) 2 + b ( − 1) + c = a − b + c = 1 p (2) = a (2) 2 + b (2) + c = 4 a + 2 b + c = 7. (b) a = 5, b = − 3, c = − 7. Section 1.2, p. 19 2. (a) A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0 1 0 0 1 1 0 1 1 1 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (b) A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0 1 1 1 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 1 0 0 0 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ . 4. a = 3, b = 1, c = 8, d = − 2. 6. (a) C + E = E + C = ⎡ ⎣ 5 − 5 8 4 2 9 5 3 4 ⎤ ⎦ . (b) Impossible. (c) [ 7 − 7 0 1 ] . (d) ⎡ ⎣ − 9 3 − 9 − 12 − 3 − 15 − 6 − 3 − 9 ⎤ ⎦ . (e) ⎡ ⎣ 0 10 − 9 8 − 1 − 2 − 5 − 4 3 ⎤ ⎦ . (f) Impossible. 8. (a) A T = ⎡ ⎣ 1 2 2 1 3 4 ⎤ ⎦ , ( A T ) T = [ 1 2 3 2 1 4 ] . (b) ⎡ ⎣ 5 4 5 − 5 2 3 8 9 4 ⎤ ⎦ . (c) [ − 6 10 11 17 ] .

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    Section 1.3 3 (d) [ 0 − 4 4 0 ] . (e) ⎡ ⎣ 3 4 6 3 9 10 ⎤ ⎦ . (f) [ 17 2 − 16 6 ] . 10. Yes: 2 [ 1 0 0 1 ] + 1 [ 1 0 0 0 ] = [ 3 0 0 2 ] . 12. ⎡ ⎣ λ − 1 − 2 − 3 − 6 λ + 2 − 3 − 5 − 2 λ − 4 ⎤ ⎦ . 14. Because the edges can be traversed in either direction. 16. Let x = ⎡ ⎢ ⎢ ⎢ ⎣ x 1 x 2 . . . x n ⎤ ⎥ ⎥ ⎥ ⎦ be an n -vector. Then x + 0 = ⎡ ⎢ ⎢ ⎢ ⎣ x 1 x 2 . . . x n ⎤ ⎥ ⎥ ⎥ ⎦ + ⎡ ⎢ ⎢ ⎢ ⎣ 0 0 . . . 0 ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ x 1 + 0 x 2 + 0 . . . x n + 0 ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎣ x 1 x 2 . . . x n ⎤ ⎥ ⎥ ⎥ ⎦ = x . 18. n ∑ i =1 m ∑ j =1 a ij = ( a 11 + a 12 + · · · + a 1 m ) + ( a 21 + a 22 + · · · + a 2 m ) + · · · + ( a n 1 + a n 2 + · · · + a nm ) = ( a 11 + a 21 + · · · + a n 1 ) + ( a 12 + a 22 + · · · + a n 2 ) + · · · + ( a 1 m + a 2 m + · · · + a nm ) = m ∑ j =1 n ∑ i =1 a ij . 19. (a) True. n ∑ i =1 ( a i + 1) = n ∑ i =1 a i + n ∑ i =1 1 = n ∑ i =1 a i + n . (b) True. n ∑ i =1 ⎛ ⎝ m ∑ j =1 1 ⎞ ⎠ = n ∑ i =1 m = mn . (c) True. ⎡ ⎣ n ∑ i =1 a i ⎤ ⎦ ⎡ ⎣ m ∑ j =1 b j ⎤ ⎦ = a 1 m ∑ j =1 b j + a 2 m ∑ j =1 b j + · · · + a n m ∑ j =1 b j = ( a 1 + a 2 + · · · + a n ) m ∑ j =1 b j = n ∑ i =1 a i m ∑ j =1 b j = m ∑ j =1 ( n ∑ i =1 a i b j ) 20. “new salaries” = u + . 08 u = 1 . 08 u . Section 1.3, p. 30 2. (a) 4. (b) 0. (c) 1. (d) 1. 4. x = 5.

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    4 Chapter 1 6. x = ± √ 2, y = ± 3. 8. x = ± 5. 10. x = 6 5 , y = 12 5 . 12. (a) Impossible. (b) ⎡ ⎣ 0 − 1 1 12 5 17 19 0 22 ⎤ ⎦ . (c) ⎡ ⎣ 15 − 7 14 23 − 5 29 13 − 1 17 ⎤ ⎦ . (d) ⎡ ⎣ 8 8 14 13 13 9 ⎤ ⎦ . (e) Impossible. 14. (a) [ 58 12 66 13 ] . (b) Same as (a). (c) [ 28 8 38 34 4 41 ] . (d) Same as (c). (e) [ 28 32 16 18 ] ; same. (f) [ − 16 − 8 − 26 − 30 0 − 31 ] . 16. (a) 1. (b) − 6. (c) [ − 3 0 1 ] . (d) ⎡ ⎣ − 1 4 2 − 2 8 4 3 − 12 − 6 ⎤ ⎦ . (e) 10. (f) ⎡ ⎣ 9 0 − 3 0 0 0 − 3 0 1 ⎤ ⎦ . (g) Impossible. 18. DI 2 = I 2 D = D . 20. [ 0 0 0 0 ] . 22. (a) ⎡ ⎢ ⎢ ⎣ 1 14 0 13 ⎤ ⎥ ⎥ ⎦ . (b) ⎡ ⎢ ⎢ ⎣ 0 18 3 13 ⎤ ⎥ ⎥ ⎦ . 24. col 1 ( AB ) = 1 ⎡ ⎣ 1 2 3 ⎤ ⎦ + 3 ⎡ ⎣ − 2 4 0 ⎤ ⎦ + 2 ⎡ ⎣ − 1 3 − 2 ⎤ ⎦ ; col 2 ( AB ) = − 1 ⎡ ⎣ 1 2 3 ⎤ ⎦ + 2 ⎡ ⎣ − 2 4 0 ⎤ ⎦ + 4 ⎡ ⎣ − 1 3 − 2 ⎤ ⎦ . 26. (a) − 5. (b) BA T 28. Let A = [ a ij ] be m × p and B = [ b ij ] be p × n . (a) Let the i th row of A consist entirely of zeros, so that a ik = 0 for k = 1 , 2 , . . . , p . Then the ( i, j ) entry in AB is p ∑ k =1 a ik b kj = 0 for j = 1 , 2 , . . . , n . (b) Let the j th column of A consist entirely of zeros, so that a kj = 0 for k = 1 , 2 , . . . , m . Then the ( i, j ) entry in BA is m ∑ k =1 b ik a kj = 0 for i = 1 , 2 , . . . , m . 30. (a) ⎡ ⎢ ⎢ ⎣ 2 3 − 3 1 1 3 0 2 0 3 2 3 0 − 4 0 0 0 1 1 1 ⎤ ⎥ ⎥ ⎦ . (b) ⎡ ⎢ ⎢ ⎣ 2 3 − 3 1 1 3 0 2 0 3 2 3 0 − 4 0 0 0 1 1 1 ⎤ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ x 1 x 2 x 3 x 4 x 5 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎣ 7 − 2 3 5 ⎤ ⎥ ⎥ ⎦ .

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    Section 1.3 5 (c) ⎡ ⎢ ⎢ ⎣ 2 3 − 3 1 1 7 3 0 2 0 3 − 2 2 3 0 − 4 0 3 0 0 1 1 1 5 ⎤ ⎥ ⎥ ⎦ 32. [ − 2 3 1 − 5 ] [ x 1 x 2 ] = [ 5 4 ] . 34. (a) 2 x 1 + x 2 + 3 x 3 + 4 x 4 = 0 3 x 1 − x 2 + 2 x 3 = 3 − 2 x 1 + x 2 − 4 x 3 + 3 x 4 = 2 (b) same as (a). 36. (a) x 1 [ 3 1 ] + x 2 [ 2 − 1 ] + x 3 [ 1 4 ] = [ 4 − 2 ] . (b) x 1 ⎡ ⎣ − 1 2 3 ⎤ ⎦ + x 2 ⎡ ⎣ 1 − 1 1 ⎤ ⎦ = ⎡ ⎣ 3 − 2 1 ⎤ ⎦ . 38. (a) [ 1 2 0 2 5 3 ] ⎡ ⎣ x 1 x 2 x 3 ⎤ ⎦ = [ 1 1 ] . (b) ⎡ ⎣ 1 2 1 1 1 2 2 0 2 ⎤ ⎦ ⎡ ⎣ x 1 x 2 x 3 ⎤ ⎦ = ⎡ ⎣ 0 0 0 ⎤ ⎦ . 39. We have u · v = n ∑ i =1 u i v i = [ u 1 u 2 · · · u n ] ⎡ ⎢ ⎢ ⎢ ⎣ v 1 v 2 . . . v n ⎤ ⎥ ⎥ ⎥ ⎦ = u T v . 40. Possible answer: ⎡ ⎣ 1 0 0 2 0 0 3 0 0 ⎤ ⎦ . 42. (a) Can say nothing. (b) Can say nothing. 43. (a) Tr( cA ) = n ∑ i =1 ca ii = c n ∑ i =1 a ii = c Tr( A ). (b) Tr( A + B ) = n ∑ i =1 ( a ii + b ii ) = n ∑ i =1 a ii + n ∑ i =1 b ii = Tr( A ) + Tr( B ). (c) Let AB = C = [ c ij ] . Then Tr( AB ) = Tr( C ) = n ∑ i =1 c ii = n ∑ i =1 n ∑ k =1 a ik b ki = n ∑ k =1 n ∑ i =1 b ki a ik = Tr( BA ) . (d) Since a T ii = a ii , Tr( A T ) = n ∑ i =1 a T ii = n ∑ i =1 a ii = Tr( A ). (e) Let A T A = B = [ b ij ] . Then b ii = n ∑ j =1 a T ij a ji = n ∑ j =1 a 2 ji = ⇒ Tr( B ) = Tr( A T A ) = n ∑ i =1 b ii = n ∑ i =1 n ∑ j =1 a 2 ij ≥ 0 . Hence, Tr( A T A ) ≥ 0.

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    6 Chapter 1 44. (a) 4. (b) 1. (c) 3. 45. We have Tr( AB − BA ) = Tr( AB ) − Tr( BA ) = 0, while Tr ([ 1 0 0 1 ]) = 2. 46. (a) Let A = [ a ij ] and B = [ b ij ] be m × n and n × p , respectively. Then b j = ⎡ ⎢ ⎢ ⎢ ⎣ b 1 j b 2 j . . . b nj ⎤ ⎥ ⎥ ⎥ ⎦ and the i th entry of A b j is n ∑ k =1 a ik b kj , which is exactly the ( i, j ) entry of AB . (b) The i th row of AB is [∑ k a ik b k 1 ∑ k a ik b k 2 · · · ∑ k a ik b kn ] . Since a i = [ a i 1 a i 2 · · · a in ] , we have a i b = [∑ k a ik b k 1 ∑ k a ik b k 2 · · · ∑ k a ik b kn ] . This is the same as the i th row of A b . 47. Let A = [ a ij ] and B = [ b ij ] be m × n and n × p , respectively. Then the j th column of AB is ( AB ) j = ⎡ ⎢ ⎣ a 11 b 1 j + · · · + a 1 n b nj . . . a m 1 b 1 j + · · · + a mn b nj ⎤ ⎥ ⎦ = b 1 j ⎡ ⎢ ⎣ a 11 . . . a m 1 ⎤ ⎥ ⎦ + · · · + b nj ⎡ ⎢ ⎣ a 1 n . . . a mn ⎤ ⎥ ⎦ = b 1 j Col 1 ( A ) + · · · + b nj Col n ( A ) . Thus the j th column of AB is a linear combination of the columns of A with coefficients the entries in b j . 48. The value of the inventory of the four types of items. 50. (a) row 1 ( A ) · col 1 ( B ) = 80(20) + 120(10) = 2800 grams of protein consumed daily by the males. (b) row 2 ( A ) · col 2 ( B ) = 100(20) + 200(20) = 6000 grams of fat consumed daily by the females. 51. (a) No. If x = ( x 1 , x 2 , . . . , x n ), then x · x = x 2 1 + x 2 2 + · · · + x 2 n ≥ 0. (b) x = 0 . 52. Let a = ( a 1 , a 2 , . . . , a n ), b = ( b 1 , b 2 , . . . , b n ), and c = ( c 1 , c 2 , . . . , c n ). Then (a) a · b = n ∑ i =1 a i b i and b · a = n ∑ i =1 b i a i , so a · b = b · a . (b) ( a + b ) · c = n ∑ i =1 ( a i + b i ) c i = n ∑ i =1 a i c i + n ∑ i =1 b i c i = a · c + b · c . (c) ( k a ) · b = n ∑ i =1 ( ka i ) b i = k n ∑ i =1 a i b i = k ( a · b ).

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    Section 1.4 7 53. The i , i th element of the matrix AA T is n ∑ k =1 a ik a T ki = n ∑ k =1 a ik a ik = n ∑ k =1 ( a ik ) 2 . Thus if AA T = O , then each sum of squares n ∑ k =1 ( a ik ) 2 equals zero, which implies a ik = 0 for each i and k . Thus A = O . 54. AC = [ 17 2 22 18 3 23 ] . CA cannot be computed. 55. B T B will be 6 × 6 while BB T is 1 × 1. Section 1.4, p. 40 1. Let A = [ a ij ] , B = [ b ij ] , C = [ c ij ] . Then the ( i, j ) entry of A + ( B + C ) is a ij + ( b ij + c ij ) and that of ( A + B ) + C is ( a ij + b ij ) + c ij . By the associative law for addition of real numbers, these two entries are equal. 2. For A = [ a ij ] , let B = [ − a ij ] . 4. Let A = [ a ij ] , B = [ b ij ] , C = [ c ij ] . Then the ( i, j ) entry of ( A + B ) C is n ∑ k =1 ( a ik + b ik ) c kj and that of AC + BC is n ∑ k =1 a ik c kj + n ∑ k =1 b ik c kj . By the distributive and additive associative laws for real numbers, these two expressions for the ( i, j ) entry are equal. 6. Let A = [ a ij ] , where a ii = k and a ij = 0 if i  = j , and let B = [ b ij ] . Then, if i  = j , the ( i, j ) entry of AB is n ∑ s =1 a is b sj = kb ij , while if i = j , the ( i, i ) entry of AB is n ∑ s =1 a is b si = kb ii . Therefore AB = kB . 7. Let A = [ a ij ] and C = [ c 1 c 2 · · · c m ] . Then CA is a 1 × n matrix whose i th entry is n ∑ j =1 c j a ij . Since A j = ⎡ ⎢ ⎢ ⎢ ⎣ a 1 j a 2 j . . . a mj ⎤ ⎥ ⎥ ⎥ ⎦ , the i th entry of n ∑ j =1 c j A j is m ∑ j =1 c j a ij . 8. (a) [ cos 2 θ sin 2 θ − sin 2 θ cos 2 θ ] . (b) [ cos 3 θ sin 3 θ − sin 3 θ cos 3 θ ] . (c) [ cos kθ sin kθ − sin kθ cos kθ ] . (d) The result is true for p = 2 and 3 as shown in parts (a) and (b). Assume that it is true for p = k . Then A k +1 = A k A = [ cos kθ sin kθ − sin kθ cos kθ ] [ cos θ sin θ − sin θ cos θ ] = [ cos kθ cos θ − sin kθ sin θ cos kθ sin θ + sin kθ cos θ − sin kθ cos θ − cos kθ sin θ cos kθ cos θ − sin kθ sin θ ] = [ cos( k + 1) θ sin( k + 1) θ − sin( k + 1) θ cos( k + 1) θ ] . Hence, it is true for all positive integers k .

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    8 Chapter 1 10. Possible answers: A = [ 1 0 0 1 ] ; A = [ 0 1 1 0 ] ; A = ⎡ ⎣ 1 √ 2 1 √ 2 1 √ 2 − 1 √ 2 ⎤ ⎦ . 12. Possible answers: A = [ 1 1 − 1 − 1 ] ; A = [ 0 0 0 0 ] ; A = [ 0 1 0 0 ] . 13. Let A = [ a ij ] . The ( i, j ) entry of r ( sA ) is r ( sa ij ), which equals ( rs ) a ij and s ( ra ij ). 14. Let A = [ a ij ] . The ( i, j ) entry of ( r + s ) A is ( r + s ) a ij , which equals ra ij + sa ij , the ( i, j ) entry of rA + sA . 16. Let A = [ a ij ] , and B = [ b ij ] . Then r ( a ij + b ij ) = ra ij + rb ij . 18. Let A = [ a ij ] and B = [ b ij ] . The ( i, j ) entry of A ( rB ) is n ∑ k =1 a ik ( rb kj ), which equals r n ∑ k =1 a ik b kj , the ( i, j ) entry of r ( AB ). 20. 1 6 A , k = 1 6 . 22. 3. 24. If A x = r x and y = s x , then A y = A ( s x ) = s ( A x ) = s ( r x ) = r ( s x ) = r y . 26. The ( i, j ) entry of ( A T ) T is the ( j, i ) entry of A T , which is the ( i, j ) entry of A . 27. (b) The ( i, j ) entry of ( A + B ) T is the ( j, i ) entry of [ a ij + b ij ] , which is to say, a ji + b ji . (d) Let A = [ a ij ] and let b ij = a ji . Then the ( i, j ) entry of ( cA ) T is the ( j, i ) entry of [ ca ij ] , which is to say, cb ij . 28. ( A + B ) T = ⎡ ⎣ 5 0 5 2 1 2 ⎤ ⎦ , ( rA ) T = ⎡ ⎣ − 4 − 8 − 12 − 4 − 8 12 ⎤ ⎦ . 30. (a) ⎡ ⎣ − 34 17 − 51 ⎤ ⎦ . (b) ⎡ ⎣ − 34 17 − 51 ⎤ ⎦ . (c) B T C is a real number (a 1 × 1 matrix). 32. Possible answers: A = [ 1 − 3 0 0 ] ; B = [ 1 2 2 3 1 ] ; C = [ − 1 2 0 1 ] . A = [ 2 0 3 0 ] ; B = [ 0 0 1 0 ] ; C = [ 0 0 0 1 ] . 33. The ( i, j ) entry of cA is ca ij , which is 0 for all i and j only if c = 0 or a ij = 0 for all i and j . 34. Let A = [ a b c d ] be such that AB = BA for any 2 × 2 matrix B . Then in particular, [ a b c d ] [ 1 0 0 0 ] = [ 1 0 0 0 ] [ a b c d ] [ a 0 c 0 ] = [ a b 0 0 ] so b = c = 0, A = [ a 0 0 d ] .

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    Section 1.5 9 Also [ a 0 0 d ] [ 1 1 0 0 ] = [ 1 1 0 0 ] [ a 0 0 d ] [ a a 0 0 ] = [ a d 0 0 ] , which implies that a = d . Thus A = [ a 0 0 a ] for some number a . 35. We have ( A − B ) T = ( A + ( − 1) B ) T = A T + (( − 1) B ) T = A T + ( − 1) B T = A T − B T by Theorem 1.4(d)). 36. (a) A ( x 1 + x 2 ) = A x 1 + A x 2 = 0 + 0 = 0 . (b) A ( x 1 − x 2 ) = A x 1 − A x 2 = 0 − 0 = 0 . (c) A ( r x 1 ) = r ( A x 1 ) = r 0 = 0 . (d) A ( r x 1 + s x 2 ) = r ( A x 1 ) + s ( A x 2 ) = r 0 + s 0 = 0 . 37. We verify that x 3 is also a solution: A x 3 = A ( r x 1 + s x 2 ) = rA x 1 + sA x 2 = r b + s b = ( r + s ) b = b . 38. If A x 1 = b and A x 2 = b , then A ( x 1 − x 2 ) = A x 1 − A x 2 = b − b = 0 . Section 1.5, p. 52 1. (a) Let I m = [ d ij ] so d ij = 1 if i = j and 0 otherwise. Then the ( i, j ) entry of I m A is m ∑ k =1 d ik a kj = d ii a ij (since all other d ’s = 0) = a ij (since d ii = 1). 2. We prove that the product of two upper triangular matrices is upper triangular: Let A = [ a ij ] with a ij = 0 for i > j ; let B = [ b ij ] with b ij = 0 for i > j . Then AB = [ c ij ] where c ij = n ∑ k =1 a ik b kj . For i > j , and each 1 ≤ k ≤ n , either i > k (and so a ik = 0) or else k ≥ i > j (so b kj = 0). Thus every term in the sum for c ij is 0 and so c ij = 0. Hence [ c ij ] is upper triangular. 3. Let A = [ a ij ] and B = [ b ij ] , where both a ij = 0 and b ij = 0 if i  = j . Then if AB = C = [ c ij ] , we have c ij = n ∑ k =1 a ik b kj = 0 if i  = j . 4. A + B = ⎡ ⎣ 9 − 1 1 0 − 2 7 0 0 3 ⎤ ⎦ and AB = ⎡ ⎣ 18 − 5 11 0 − 8 − 7 0 0 0 ⎤ ⎦ . 5. All diagonal matrices.

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    10 Chapter 1 6. (a) [ 7 − 2 − 3 10 ] (b) [ − 9 − 11 22 13 ] (c) [ 20 − 20 4 76 ] 8. A p A q = ( A · A · · · A ) ︸ ︷︷ ︸ p factors ( A · A · · · A ) ︸ ︷︷ ︸ q factors ︸ ︷︷ ︸ p + q factors = A p + q ; ( A p ) q = A p A p A p · · · A p ︸ ︷︷ ︸ q factors = A q summands ︷ ︸︸ ︷ p + p + · · · + p = A pq . 9. We are given that AB = BA . For p = 2, ( AB ) 2 = ( AB )( AB ) = A ( BA ) B = A ( AB ) B = A 2 B 2 . Assume that for p = k , ( AB ) k = A k B k . Then ( AB ) k +1 = ( AB ) k ( AB ) = A k B k · A · B = A k ( B k − 1 AB ) B = A k ( B k − 2 AB 2 ) B = · · · = A k +1 B k +1 . Thus the result is true for p = k + 1. Hence it is true for all positive integers p . For p = 0, ( AB ) 0 = I n = A 0 B 0 . 10. For p = 0, ( cA ) 0 = I n = 1 · I n = c 0 · A 0 . For p = 1, cA = cA . Assume the result is true for p = k : ( cA ) k = c k A k , then for k + 1: ( cA ) k +1 = ( cA ) k ( cA ) = c k A k · cA = c k ( A k c ) A = c k ( cA k ) A = ( c k c )( A k A ) = c k +1 A k +1 . 11. True for p = 0: ( A T ) 0 = I n = I T n = ( A 0 ) T . Assume true for p = n . Then ( A T ) n +1 = ( A T ) n A T = ( A n ) T A T = ( AA n ) T = ( A n +1 ) T . 12. True for p = 0: ( A 0 ) − 1 = I − 1 n = I n . Assume true for p = n . Then ( A n +1 ) − 1 = ( A n A ) − 1 = A − 1 ( A n ) − 1 = A − 1 ( A − 1 ) n = ( A − 1 ) n +1 . 13. ( 1 k A − 1 ) ( kA ) = ( 1 k · k ) A − 1 A = I n and ( kA ) ( 1 k A − 1 ) = ( k · 1 k ) AA − 1 = I n . Hence, ( kA ) − 1 = 1 k A − 1 for k  = 0. 14. (a) Let A = kI n . Then A T = ( kI n ) T = kI T n = kI n = A . (b) If k = 0, then A = kI n = 0 I n = O , which is singular. If k  = 0, then A − 1 = ( kA ) − 1 = 1 k A − 1 , so A is nonsingular. (c) No, the entries on the main diagonal do not have to be the same. 16. Possible answers: [ a b 0 a ] . Infinitely many. 17. The result is false. Let A = [ 1 2 3 4 ] . Then AA T = [ 5 11 11 25 ] and A T A = [ 10 14 14 20 ] . 18. (a) A is symmetric if and only if A T = A , or if and only if a ij = a T ij = a ji . (b) A is skew symmetric if and only if A T = − A , or if and only if a T ij = a ji = − a ij . (c) a ii = − a ii , so a ii = 0. 19. Since A is symmetric, A T = A and so ( A T ) T = A T . 20. The zero matrix. 21. ( AA T ) T = ( A T ) T A T = AA T . 22. (a) ( A + A T ) T = A T + ( A T ) T = A T + A = A + A T .

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    Section 1.5 11 (b) ( A − A T ) T = A T − ( A T ) T = A T − A = − ( A − A T ). 23. ( A k ) T = ( A T ) k = A k . 24. (a) ( A + B ) T = A T + B T = A + B . (b) If AB is symmetric, then ( AB ) T = AB , but ( AB ) T = B T A T = BA , so AB = BA . Conversely, if AB = BA , then ( AB ) T = B T A T = BA = AB , so AB is symmetric. 25. (a) Let A = [ a ij ] be upper triangular, so that a ij = 0 for i > j . Since A T = [ a T ij ] , where a T ij = a ji , we have a T ij = 0 for j > i , or a T ij = 0 for i < j . Hence A T is lower triangular. (b) Proof is similar to that for (a). 26. Skew symmetric. To show this, let A be a skew symmetric matrix. Then A T = − A . Therefore ( A T ) T = A = − A T . Hence A T is skew symmetric. 27. If A is skew symmetric, A T = − A . Thus a ii = − a ii , so a ii = 0. 28. Suppose that A is skew symmetric, so A T = − A . Then ( A k ) T = ( A T ) k = ( − A ) k = − A k if k is a positive odd integer, so A k is skew symmetric. 29. Let S = ( 1 2 ) ( A + A T ) and K = ( 1 2 ) ( A − A T ). Then S is symmetric and K is skew symmetric, by Exercise 18. Thus S + K = ( 1 2 ) ( A + A T + A − A T ) = ( 1 2 ) (2 A ) = A. Conversely, suppose A = S + K is any decomposition of A into the sum of a symmetric and skew symmetric matrix. Then A T = ( S + K ) T = S T + K T = S − K A + A T = ( S + K ) + ( S − K ) = 2 S, S = ( 1 2 ) ( A + A T ) , A − A T = ( S + K ) − ( S − K ) = 2 K, K = ( 1 2 ) ( A − A T ) 30. S = 1 2 ⎡ ⎣ 2 7 3 7 12 3 3 3 6 ⎤ ⎦ and K = 1 2 ⎡ ⎣ 0 − 1 − 7 1 0 1 7 − 1 0 ⎤ ⎦ . 31. Form [ 2 3 4 6 ] [ w x y z ] = [ 1 0 0 1 ] . Since the linear systems 2 w + 3 y = 1 4 w + 6 y = 0 and 2 x + 3 z = 0 4 x + 6 z = 1 have no solutions, we conclude that the given matrix is singular. 32. D − 1 = ⎡ ⎢ ⎢ ⎢ ⎣ 1 4 0 0 0 − 1 2 0 0 0 1 3 ⎤ ⎥ ⎥ ⎥ ⎦ . 34. A = ⎡ ⎣ − 1 2 1 2 2 − 1 ⎤ ⎦ . 36. (a) [ 1 2 1 3 ] [ 4 6 ] = [ 16 22 ] . (b) [ 38 53 ] .

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    12 Chapter 1 38. [ − 9 − 6 ] . 40. [ 8 9 ] . 42. Possible answer: [ 1 0 0 0 ] + [ 0 0 0 1 ] = [ 1 0 0 1 ] . 43. Possible answer: [ 1 2 3 4 ] + [ − 1 − 2 3 4 ] = [ 0 0 6 8 ] . 44. The conclusion of the corollary is true for r = 2, by Theorem 1.6. Suppose r ≥ 3 and that the conclusion is true for a sequence of r − 1 matrices. Then ( A 1 A 2 · · · A r ) − 1 = [( A 1 A 2 · · · A r − 1 ) A r ] − 1 = A − 1 r ( A 1 A 2 · · · A r − 1 ) − 1 = A − 1 r A − 1 r − 1 · · · A − 1 2 A − 1 1 . 45. We have A − 1 A = I n = AA − 1 and since inverses are unique, we conclude that ( A − 1 ) − 1 = A . 46. Assume that A is nonsingular, so that there exists an n × n matrix B such that AB = I n . Exercise 28 in Section 1.3 implies that AB has a row consisting entirely of zeros. Hence, we cannot have AB = I n . 47. Let A = ⎡ ⎢ ⎢ ⎢ ⎣ a 11 0 0 · · · 0 0 a 22 0 · · · 0 . . . 0 0 · · · · · · a nn ⎤ ⎥ ⎥ ⎥ ⎦ , where a ii  = 0 for i = 1 , 2 , . . . , n . Then A − 1 = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 a 11 0 0 · · · 0 0 1 a 22 0 · · · 0 . . . 0 0 · · · · · · 1 a nn ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ as can be verified by computing AA − 1 = A − 1 A = I n . 48. A 4 = ⎡ ⎣ 16 0 0 0 81 0 0 0 625 ⎤ ⎦ . 49. A p = ⎡ ⎢ ⎢ ⎢ ⎣ a p 11 0 0 · · · 0 0 a p 22 0 · · · 0 . . . 0 0 · · · · · · a p nn ⎤ ⎥ ⎥ ⎥ ⎦ . 50. Multiply both sides of the equation by A − 1 . 51. Multiply both sides by A − 1 .

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