Explanation Given: The given system is, X ′ = ( 0 0 1 0 1 0 1 0 0 ) X with initial condition X ( 0 ) = ( 1 2 5 ) . Calculation: For the given system, the coefficient matrix is A = ( 0 0 1 0 1 0 1 0 0 ) . The characteristic equation is | ( A − λ I ) | = 0 where, I is the identity matrix. Substitute the values of matrix A and I in above equation. | − λ 0 1 0 1 − λ 0 1 0 − λ | = 0 ( − λ ) ( λ ( λ − 1 ) ) + ( λ − 1 ) = 0 − [ λ 2 ( λ − 1 ) − ( λ − 1 ) ] = 0 − ( λ + 1 ) ( λ − 1 ) 2 = 0 Therefore, the eigenvalues of the above system is λ 1 = − 1 and λ 2 = 1 with multiplicity 2 . Obtain the eigenvector K 1 corresponding to the eigenvalue, λ 1 = − 1 . Solve the equation ( A − λ 1 I ) K 1 = 0 where, K 1 = ( x 1 y 1 z 1 ) . Substitute the values of A , λ 1 and K 1 in the above equation gives, [ ( 0 0 1 0 1 0 1 0 0 ) − ( − 1 0 0 0 − 1 0 0 0 − 1 ) ] ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( 1 0 1 0 2 0 1 0 1 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) By applying the row transformation R 3 → R 3 − R 1 in the above system of equations, ( 1 0 1 0 2 0 0 0 0 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( x 1 + z 1 2 y 1 0 ) = ( 0 0 0 ) Solve the above system of equations gives, y 1 = 0 and x 1 = − z 1 Assume that z 1 = p 1 for some scalar p 1 which implies, x 1 = − p 1 . Thus, the value of ( x 1 y 1 z 1 ) is, ( x 1 y 1 z 1 ) = ( − p 1 0 p 1 ) = p 1 ( − 1 0 1 ) Therefore, the eigenvector K 1 corresponding to λ 1 = − 1 is ( − 1 0 1 ) . Obtain the eigenvector corresponding to λ 2 = 1 . Solve the equation ( A − λ 2 I ) K 2 = 0 where K 2 is given by ( x 2 y 2 z 2 ) . Substitute the values of A , λ 2 and K 2 in the above equation gives, [ ( 0 0 1 0 1 0 1 0 0 ) − ( 1 0 0 0 1 0 0 0 1 ) ] ( x 2 y 2 z 2 ) = ( 0 0 0 ) ( − 1 0 1 0 0 0 1 0 − 1 ) ( x 2 y 2 z 2 ) = ( 0 0 0 ) By applying the row transformation R 3 → R 3 + R 1 in the above system of equations, ( − 1 0 1 0 0 0 0 0 0 ) ⋅ ( x 2 y 2 z 2 ) = ( 0 0 0 ) ( − x 2 + z 2 0 0 ) = ( 0 0 0 ) Solve the above system of equation gives x 2 = z 2

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ISBN: 9781337652469

Author: ZILL

Publisher: CENGAGE L

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Solution Summary: The author explains how to solve the given initial value problem. The coefficient matrix is A and I.

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