Explanation Given: The given system is, d x d t = 3 x − y − z d y d t = x + y − z d z d t = x − y + z Calculation: For the given system, the coefficient matrix is A = ( 3 − 1 − 1 1 1 − 1 1 − 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. | 3 − λ − 1 − 1 1 1 − λ − 1 1 − 1 1 − λ | = 0 ( 3 − λ ) [ ( 1 − λ ) 2 − 1 ] + 1 [ 1 − λ + 1 ] − 1 [ − 1 + λ − 1 ] = 0 ( 3 − λ ) [ λ 2 − 2 λ ] + ( 2 − λ ) + ( 2 − λ ) = 0 ( 3 − λ ) ( λ − 2 ) λ + ( 2 − λ ) + ( 2 − λ ) = 0 Simplify further, ( λ − 2 ) [ ( 3 − λ ) λ − 1 − 1 ] = 0 ( λ − 2 ) ( 3 λ − λ 2 − 2 ) = 0 ( λ − 2 ) ( λ − 2 ) ( 1 − λ ) = 0 ( 1 − λ ) ( λ − 2 ) 2 = 0 Therefore, the eigenvalues of the above system is λ 1 = 1 with multiplicity 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, [ ( 3 − 1 − 1 1 1 − 1 1 − 1 1 ) − ( 1 0 0 0 1 0 0 0 1 ) ] ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( 2 − 1 − 1 1 0 − 1 1 − 1 0 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) By applying the row transformation R 3 → R 3 − R 2 in the above system of equations, ( 2 − 1 − 1 1 0 − 1 0 − 1 1 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Now, by row transformation R 1 → R 1 − 2 R 2 . ( 0 − 1 1 1 0 − 1 0 − 1 1 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Which on applying R 1 → R 1 − R 3 in above system, ( 0 0 0 1 0 − 1 0 − 1 1 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( 0 x 1 − z 1 − y 1 + z 1 ) = ( 0 0 0 ) Solve the above system of equation gives, x 1 = z 1 and y 1 = z 1 Assume that z 1 = p 1 for some scalar p 1 , which implies, x 1 = p 1 , y 1 = p 1 Thus, the value of ( x 1 y 1 z 1 ) is, ( x 1 y 1 z 1 ) = ( p 1 p 1 p 1 ) = p 1 ( 1 1 1 ) Therefore, the eigenvector K 1 corresponding to λ 1 = 1 is ( 1 1 1 )

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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The general solution of the given system.

Solution Summary: The author explains that the general solution of the given system is X(t)=c_1 left, where I is the identity matrix.

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