Explanation Formula used: The eigen values of a given matrix A is obtained by solving the equation det ( A − λ I ) = 0 where I is an n × n matrix. Calculation: It is given that x ′ = ( 1 1 2 1 2 1 2 1 1 ) x . Let A = ( 1 1 2 1 2 1 2 1 1 ) . Substitute A = ( 1 1 2 1 2 1 2 1 1 ) in det ( A − λ I ) = 0 and obtain the eigen values as shown below: | [ 1 1 2 1 2 1 2 1 1 ] − λ [ 1 0 0 0 1 0 0 0 1 ] | = 0 | 1 − λ 1 2 1 2 − λ 1 2 1 1 − λ | = 0 ⋯ ⋯ ( 1 ) ( 1 − λ ) | 2 − λ 1 2 1 − λ | − | 1 1 2 1 − λ | + 2 | 1 2 − λ 2 1 | = 0 λ 3 − 4 λ 2 − λ + 4 = 0 λ = 1 , − 1 , 4 Thus, λ = 1 , − 1 , 4 . Substitute λ = 1 in equation (1) and obtain the eigen vector as follows: [ 1 − λ 1 2 1 2 − λ 1 2 1 1 − λ ] = [ 1 − 1 1 2 1 2 − 1 1 2 1 1 − 1 ] = [ 0 1 2 1 1 1 2 1 0 ] From the first row of the above matrix, the corresponding eigen vector associated with λ = 1 is obtained as v 1 = [ 1 − 2 1 ] . Substitute λ 2 = − 1 in equation (1) and obtain the eigen vector as follows: [ 1 − λ 1 2 1 2 − λ 1 2 1 1 − λ ] = [ 1 − ( − 1 ) 1 2 1 2 − ( − 1 ) 1 2 1 1 − ( − 1 ) ] = [ 2 1 2 1 3 1 2 1 2 ] ∼ [ 1 0 1 0 1 0 0 0 0 ] From the first row of the above matrix, the corresponding eigen vector associated with λ 2 = − 1 is obtained as v 2 = [ − 1 0 1 ]

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 7.5, Problem 11P

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Elementary Differential Equations

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