Explanation Given: The matrix A is given as, A = [ − 7 − 4 18 11 ] Approach: An n × n matrix that is similar to a diagonal matrix is said to be diagonalizable. An n × n matrix A is similar to a diagonal matrix if and only if A is nondefective. In such a case, if v 1 , v 2 , ......... , v n denote n linearly independent eigenvectors of A and S = [ v 1 , v 2 , .......... , v n ] , Then, S − 1 A S = diag ( λ 1 , λ 2 , ......... , λ n ) = D , Where D is a diagonal matrix and λ 1 , λ 2 , ........ , λ n are the eigenvalues of A (non necessarily distinct) corresponding in the eigenvectors v 1 , v 2 , ............ , v n which forms the diagonal elements of the matrix D . For every positive integer k , A k = S D k S − 1 . And, D k = diag ( λ 1 k , λ 2 k , ....... , λ n k ) Calculation: Consider the given matrix A , A = [ − 7 − 4 18 11 ] The characteristic equation of A is given as det ( A − λ I ) = 0 , det ( A − λ I ) = 0 | [ − 7 − 4 18 11 ] − [ λ 0 0 λ ] | = 0 | − 7 − λ − 4 18 11 − λ | = 0 Solve the above equation for the eigenvalues ( λ ) of A . | − 7 − λ − 4 18 11 − λ | = 0 ( − 7 − λ ) ( 11 − λ ) − ( 18 ) ( − 4 ) = 0 − 77 + 7 λ − 11 λ + λ 2 + 72 = 0 λ 2 − 4 λ − 5 = 0 Further solve the above equation. λ 2 − 5 λ + λ − 5 = 0 λ ( λ − 5 ) + 1 ( λ − 5 ) = 0 ( λ + 1 ) ( λ − 5 ) = 0 λ = − 1 , 5 Therefore, the eigenvalue of A are − 1 and 5 . The eigenvector corresponding to the eigenvalue λ 1 = − 1 is determined based on the linear system given as, ( A − λ I ) v = 0 ( [ − 7 − 4 18 11 ] − [ λ 0 0 λ ] ) [ v 1 v 2 ] = [ 0 0 ] [ − 7 − λ − 4 18 11 − λ ] [ v 1 v 2 ] = [ 0 0 ] Substitute − 1 for λ in the above equation. [ − 7 − ( − 1 ) − 4 18 11 − ( − 1 ) ] [ v 1 v 2 ] = [ 0 0 ] [ − 7 + 1 − 4 18 11 + 1 ] [ v 1 v 2 ] = [ 0 0 ] [ − 6 − 4 18 12 ] [ v 1 v 2 ] = [ 0 0 ] The augmented matrix of the above expression is, [ − 6 − 4 18 12 | 0 0 ] Apply R 2 → R 2 + 3 R 1 . [ − 6 − 4 0 0 | 0 0 ] Apply R 1 → R 1 ( − 1 2 ) . [ 3 2 0 0 | 0 0 ] Thus, [ 3 2 0 0 ] [ v 1 v 2 ] = [ 0 0 ] [ 3 v 1 + 2 v 2 0 ] = [ 0 0 ] 3 v 1 + 2 v 2 = 0 Solve the above equation for v 1 . v 1 = − 2 3 v 2 For, v 2 = 3 k , then v 1 = − 2 k . Thus, v 1 = − 2 k and v 2 = 3 k , the general solution to this system is, v = [ v 1 , v 2 ] T = [ − 2 k , 3 k ] T = [ − 2 k 3 k ] = k [ − 2 3 ] : k ∈ ℝ Here, k is a scalar. The eigenvector corresponding to the eigenvalue λ 1 = 5 is determined based on the linear system given as, ( A − λ I ) v = 0 ( [ − 7 − 4 18 11 ] − [ λ 0 0 λ ] ) [ v 1 v 2 ] = [ 0 0 ] [ − 7 − λ − 4 18 11 − λ ] [ v 1 v 2 ] = [ 0 0 ] Substitute 5 for λ in the above equation. [ − 7 − ( 5 ) − 4 18 11 − ( 5 ) ] [ v 1 v 2 ] = [ 0 0 ] [ − 7 − 5 − 4 18 11 − 5 ] [ v 1 v 2 ] = [ 0 0 ] [ − 12 − 4 18 6 ] [ v 1 v 2 ] = [ 0 0 ] The augmented matrix of the above expression is, [ − 12 − 4 18 6 | 0 0 ] Apply R 1 → − 1 4 R 1 and R 2 → 1 6 R 2

Differential Equations and Linear Algebra (4th Edition)

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ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 7.3, Problem 27P

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The value of A3 and A5 for the matrix A based on the preceding two problems.

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Chapter 7.3, Problem 26P

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Kate, Luke, Mary and Nancy are sharing a cake. The cake had previously been divided into four slices (s1, s2, s3 and s4). The following table shows the values of the slices in the eyes of each player. What is fair share to nancy? S1 S2 S3 S4 Kate $4.00 $6.00 $6.00 $4.00 Luke $5.30 $5.00 $5.25 $5.45 Mary $4.25 $4.50 $3.50 $3.75 Nancy $6.00 $4.00 $4.00 $6.00

Kate, Luke, Mary and Nancy are sharing a cake. The cake had previously been divided into four slices (s1, s2, s3 and s4). The following table shows the values of the slices in the eyes of each player. S1 S2 S3 S4 Kate $4.00 $6.00 $6.00 $4.00 Luke $5.30 $5.00 $5.25 $5.45 Mary $4.25 $4.50 $3.50 $3.75 Nancy $6.00 $4.00 $4.00 $6.00 how much is the cak worth to mary

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Chapter 7 Solutions

Differential Equations and Linear Algebra (4th Edition)

Ch. 7.1 - Prob. 1P Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.1 - Prob. 4P Ch. 7.1 - Prob. 5P Ch. 7.1 - Given that v1=(2,1) and v2=(1,1) are eigenvectors...Ch. 7.1 - Prob. 7P Ch. 7.1 - Prob. 8P Ch. 7.1 - Prob. 9P Ch. 7.1 - Prob. 11P

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The value of A 3 and A 5 for the matrix A based on the preceding two problems.

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