Explanation Procedure used: A nonzero vector v is an eigenvector of the square matrix A if there is a number λ , called an eigenvalue of A , such that, ( A − λI ) v = 0 . ... ( 1 ) If A is an n × n matrix, the equation det ( A − λ I ) = 0 is called the characteristic equation of A . Calculation: Consider the matrix A = ( 0 − 3 − 3 0 − 3 − 3 0 2 2 ) . The characteristic equation of matrix A is given as follows. | A − λ I | = 0 | ( 0 − 3 − 3 0 − 3 − 3 0 2 2 ) − λ ( 1 0 0 0 1 0 0 0 1 ) | = 0 | ( − λ − 3 − 3 0 − 3 − λ − 3 0 2 2 − λ ) | = 0 ( − λ ) ⋅ | − 3 − λ − 3 2 2 − λ | = 0 Solve further. ( − λ ) ⋅ [ ( − 3 − λ ) ( 2 − λ ) + 6 ] = 0 ( − λ ) ⋅ [ − 6 + 3 λ − 2 λ + λ 2 + 6 ] = 0 − λ 2 ( λ+1 ) = 0 Thus, the eigenvalues of matrix A are λ 1 , 2 = 0 and λ 3 = − 1 . Let v 1 = [ x 1 y 1 z 1 ] T be the eigenvector corresponding to λ 1 = 0 . ( A − ( 0 ) I ) v 1 = 0 ( 0 − 3 − 3 0 − 3 − 3 0 2 2 ) v 1 = 0 Form the augmented matrix by reducing the column to zero: ( 0 − 3 − 3 0 − 3 − 3 0 2 2 ) → R 2 − R 1 , R 1 → − 1 3 R 1 , R 3 − 2 R 1 ( 0 1 1 0 0 0 0 0 0 ) So, ( 0 1 1 0 0 0 0 0 0 ) v 1 = 0 ( 0 1 1 0 0 0 0 0 0 ) [ x 1 y 1 z 1 ] = 0 The system of equations thus obtained is, x 1 = 0 y 1 + z 1 = 0 Further solve the above system

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ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Chapter 6, Problem 6RE

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The eigenvalues and corresponding eigenvectors of the matrix A=(0−3−30−3−3022).

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Chapter 6 Solutions

Introductory Differential Equations

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The eigenvalues and corresponding eigenvectors of the matrix A = ( 0 − 3 − 3 0 − 3 − 3 0 2 2 ) .

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The eigenvalues and corresponding eigenvectors of the matrix A=(0−3−30−3−3022).

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