Explanation Given: The given system is, X ′ = ( 1 0 0 0 3 1 0 − 1 1 ) X Calculation: For the given system, the coefficient matrix is A = ( 1 0 0 0 3 1 0 − 1 1 ) . The characteristic equation for the matrix A is | ( A − λ I ) | = 0 where I is the identity matrix. Substitute the values of matrix A and I in above expression. | 1 − λ 0 0 0 3 − λ 1 0 − 1 1 − λ | = 0 ( 1 − λ ) [ ( 3 − λ ) ( 1 − λ ) + 1 ] = 0 ( 1 − λ ) ( λ 2 − 4 λ + 4 ) = 0 ( 1 − λ ) ( λ − 2 ) 2 = 0 Therefore, the eigenvalues of the above system is λ 1 = 1 and λ 2 = 2 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, [ ( 1 0 0 0 3 1 0 − 1 1 ) − ( 1 0 0 0 1 0 0 0 1 ) ] ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( 0 0 0 0 2 1 0 − 1 0 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( 0 2 y 1 + z 1 − y 1 ) = ( 0 0 0 ) Which is equivalent to the equations, 2 y 1 + z 1 = 0 and y 1 = 0 . On solving further gives, z 1 = 0 . Assume that x 1 = n for some scalar n . Therefore, the value of ( x 1 y 1 z 1 ) is, ( x 1 y 1 z 1 ) = ( n 0 0 ) = n ( 1 0 0 ) Thus, the eigenvector K 1 corresponding to λ 1 = 1 is ( 1 0 0 ) . Obtain the eigenvectors K 2 corresponding to λ 2 = 2 . 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, [ ( 1 0 0 0 3 1 0 − 1 1 ) − ( 2 0 0 0 2 0 0 0 2 ) ] ( x 2 y 2 z 2 ) = ( 0 0 0 ) ( − 1 0 0 0 1 1 0 − 1 − 1 ) ( x 2 y 2 z 2 ) = ( 0 0 0 ) By applying the row transformation R 3 → R 3 + R 2 in above system, ( − 1 0 0 0 1 1 0 0 0 ) ( x 2 y 2 z 2 ) = ( 0 0 0 ) ( − x 2 y 2 + z 2 0 ) = ( 0 0 0 ) On solving further the above equation and obtain x 2 = 0 and y 2 + z 2 = 0 which gives y 2 = − z 2

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 10.2, Problem 28E

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