Explanation Given: The given matrix is: [ − 3 1 0 0 − 3 1 4 − 8 2 ] Procedure used: Finding the eigenvalues and eigenvectors: Expanding the determinant of | A − r I | = 0 and then, solve the n th–degree polynomial in r . Therefore, finding the eigenvalues is equal to the finding the zero of the polynomial. The above expression is called characteristics equation. The corresponding nontrivial solutions r are called eigenvectors and It get by solving the ( A − r I ) u = 0 Here, u = [ u 1 u 2 u 3 ] . Calculation: Consider the matrix: A = [ − 3 1 0 0 − 3 1 4 − 8 2 ] Substitute the value of matrix A in the characteristic equation and solve for r : | [ − 3 1 0 0 − 3 1 4 − 8 2 ] − r [ 1 0 0 0 1 0 0 0 1 ] | = 0 | [ − 3 1 0 0 − 3 1 4 − 8 2 ] − [ r 0 0 0 r 0 0 0 r ] | = 0 | [ − 3 − r 1 0 0 − 3 − r 1 4 − 8 2 − r ] | = 0 ( − 3 − r ) ⋅ | − 3 − r 1 − 8 2 − r | − 1 ⋅ | 0 1 4 2 − r | = 0 Further simplify as: ( − 3 − r ) ⋅ [ ( − 3 − r ) ( 2 − r ) + 8 ] + 4 = 0 ( − 3 − r ) [ − 6 + 3 r − 2 r + r 2 + 8 ] + 4 = 0 ( − 3 − r ) [ r + r 2 + 2 ] + 4 = 0 − r 3 − 4 r 2 − 5 r − 2 = 0 Further simplify as: − r 3 − 4 r 2 − 5 r − 2 = 0 − ( r + 1 ) ( r + 1 ) ( r + 2 ) = 0 Take first factor: ( r + 1 ) = 0 r = − 1 Take second factor: ( r + 2 ) = 0 r = − 2 Therefore, the eigenvalues are r 1 , 2 = − 1 and r 3 = − 2 . Find the eigenvector corresponding to eigenvalue r 1 = − 1 by using the formula ( A − r I ) u = 0 :

Fundamentals of Differential Equations [With CDROM] - 7th Edition

7th Edition

ISBN: 9780321410481

Author: Saff, Edward B., Snider, Arthur David, Nagle, R. Kent

Publisher: Addison Wesley

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Chapter 9.5, Problem 8E

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The eigenvalues and eigenvectors for the given matrix.

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Chapter 9.5, Problem 7E

Chapter 9.5, Problem 9E

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The eigenvalues and eigenvectors for the given matrix.

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To find: The eigenvalues and eigenvectors for the given matrix.

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Chapter 9 Solutions

To find:

The eigenvalues and eigenvectors for the given matrix.