Explanation Given: The given matrix A is, A = [ 1 2 3 4 − 1 − 2 − 3 − 4 2 4 6 8 − 2 − 4 − 6 − 8 ] 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 ) , Where λ 1 , λ 2 , ........ , λ n are the eigenvalues of A (non necessarily distinct) corresponding in the eigenvectors v 1 , v 2 , ............ , v n . Calculation: Consider the given matrix A , A = [ 1 2 3 4 − 1 − 2 − 3 − 4 2 4 6 8 − 2 − 4 − 6 − 8 ] The order the matrix A is n = 4 . The characteristic equation of A is given as det ( A − λ I ) = 0 , det ( A − λ I ) = 0 | [ 1 2 3 4 − 1 − 2 − 3 − 4 2 4 6 8 − 2 − 4 − 6 − 8 ] − [ λ 0 0 0 0 λ 0 0 0 0 λ 0 0 0 0 λ ] | = 0 | 1 − λ 2 3 4 − 1 − 2 − λ − 3 − 4 2 4 6 − λ 8 − 2 − 4 − 6 − 8 − λ | = 0 Solve the above equation for the eigenvalues ( λ ) of A , | 1 − λ 2 3 4 − 1 − 2 − λ − 3 − 4 2 4 6 − λ 8 − 2 − 4 − 6 − 8 − λ | = 0 λ 4 + 3 λ 3 = 0 λ 3 ( λ + 3 ) = 0 λ = 0 , − 3 Therefore, the eigenvalues of A are 0 and − 3 . The eigenvector corresponding to the eigenvalue λ 1 = 0 is determined based on the linear system given as, ( A − λ I ) v = 0 ( [ 1 2 3 4 − 1 − 2 − 3 − 4 2 4 6 8 − 2 − 4 − 6 − 8 ] − [ λ 0 0 0 0 λ 0 0 0 0 λ 0 0 0 0 λ ] ) [ v 1 v 2 v 3 v 4 ] = [ 0 0 0 0 ] [ 1 − λ 2 3 4 − 1 − 2 − λ − 3 − 4 2 4 6 − λ 8 − 2 − 4 − 6 − 8 − λ ] [ v 1 v 2 v 3 v 4 ] = [ 0 0 0 0 ] Substitute 0 for λ in the above equation. [ 1 − 0 2 3 4 − 1 − 2 − 0 − 3 − 4 2 4 6 − 0 8 − 2 − 4 − 6 − 8 − 0 ] [ v 1 v 2 v 3 v 4 ] = [ 0 0 0 0 ] [ 1 2 3 4 − 1 − 2 − 3 − 4 2 4 6 8 − 2 − 4 − 6 − 8 ] [ v 1 v 2 v 3 v 4 ] = [ 0 0 0 0 ] The augmented matrix of the above expression is, [ 1 2 3 4 − 1 − 2 − 3 − 4 2 4 6 8 − 2 − 4 − 6 − 8 | 0 0 0 0 ] Apply R 2 → R 2 + R 1 . [ 1 2 3 4 0 0 0 0 2 4 6 8 − 2 − 4 − 6 − 8 | 0 0 0 0 ] Apply R 4 → R 4 + R 3 . [ 1 2 3 4 0 0 0 0 2 4 6 8 0 0 0 0 | 0 0 0 0 ] Apply R 3 → R 3 − 2 R 1 . [ 1 2 3 4 0 0 0 0 0 0 0 0 0 0 0 0 | 0 0 0 0 ] Thus, [ 1 2 3 4 0 0 0 0 0 0 0 0 0 0 0 0 ] [ v 1 v 2 v 3 v 4 ] = [ 0 0 0 0 ] [ v 1 + 2 v 2 + 3 v 3 + 4 v 4 0 0 0 ] = [ 0 0 0 0 ] v 1 + 2 v 2 + 3 v 3 + 4 v 4 = 0 For, v 2 = r , v 3 = s and v 4 = t , the value of v 1 is, v 1 + 2 r + 3 s + 4 t = 0 v 1 = − 2 r − 3 s − 4 t Thus, v 1 = − 2 r − 3 s − 4 t , v 2 = r , v 3 = s and v 4 = t , the solution to this system is, v = [ v 1 , v 2 , v 3 , v 4 ] T = [ − 2 r − 3 s − 4 t , r , s , t ] T = [ − 2 r − 3 s − 4 t r s t ] : r , s , t ∈ ℝ = r [ − 2 1 0 0 ] + s [ − 3 0 1 0 ] + t [ − 4 0 0 1 ] : r , s , t ∈ ℝ The eigenvector corresponding to the eigenvalue λ 4 = − 3 is determined based on the linear system given as, ( A − λ I ) v = 0 ( [ 1 2 3 4 − 1 − 2 − 3 − 4 2 4 6 8 − 2 − 4 − 6 − 8 ] − [ λ 0 0 0 0 λ 0 0 0 0 λ 0 0 0 0 λ ] ) [ v 1 v 2 v 3 v 4 ] = [ 0 0 0 0 ] [ 1 − λ 2 3 4 − 1 − 2 − λ − 3 − 4 2 4 6 − λ 8 − 2 − 4 − 6 − 8 − λ ] [ v 1 v 2 v 3 v 4 ] = [ 0 0 0 0 ] Substitute − 3 for λ in the above equation

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Question

Chapter 7.3, Problem 15P

To determine

Whether the given matrix A is diagonalizable, when possible find the matrix S such that S−1AS=diag(λ1,λ2,........,λn).

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Chapter 7.3, Problem 14P

Chapter 7.3, Problem 18P

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2.8.1

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

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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Whether the given matrix A is diagonalizable, when possible find the matrix S such that S − 1 A S = diag ( λ 1 , λ 2 , ........ , λ n ) .

Whether the given matrix A is diagonalizable, when possible find the matrix S such that S−1AS=diag(λ1,λ2,........,λn).

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Chapter 7 Solutions