Explanation Given: The given system is, d x d t = 3 x + 2 y + 4 z d y d t = 2 x + 2 z d z d t = 4 x + 2 y + 3 z Calculation: For the given system, the coefficient matrix is A = ( 3 2 4 2 0 2 4 2 3 ) . 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 2 4 2 − λ 2 4 2 3 − λ | = 0 ( 3 − λ ) [ ( λ − 3 ) λ − 4 ] − 2 [ 6 − 2 λ − 8 ] + 4 [ 4 + 4 λ ] = 0 ( 3 − λ ) [ λ 2 − 3 λ − 4 ] − 2 [ − 2 − 2 λ ] + 16 ( 1 + λ ) = 0 ( 3 − λ ) ( λ − 4 ) ( λ + 1 ) + 4 ( 1 + λ ) + 16 ( 1 + λ ) = 0 Simplify further, ( λ + 1 ) [ ( 3 − λ ) ( λ − 4 ) + 4 + 16 ] = 0 ( λ + 1 ) [ − 12 + 7 λ − λ 2 + 20 ] = 0 ( λ + 1 ) ( 8 + 7 λ − λ 2 ) = 0 ( λ + 1 ) ( 8 − λ ) ( λ + 1 ) = 0 Therefore, the eigenvalues of the above system is λ 1 = 8 and λ 2 = − 1 with multiplicity 2 Obtain the eigenvector K 1 corresponding to the eigenvalue λ 1 = 8 . 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 2 4 2 0 2 4 2 3 ) − ( 8 0 0 0 8 0 0 0 8 ) ] ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( − 5 2 4 2 − 8 2 4 2 − 5 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Now replace the second Row with the first Row, ( 2 − 8 2 − 5 2 4 4 2 − 5 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) By applying the row transformation R 1 → 1 2 R 1 in the above system of equations, ( 1 − 4 1 − 5 2 4 4 2 − 5 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Again by row transformation, R 2 → R 2 + 5 R 1 ( 1 − 4 1 0 − 18 9 4 2 − 5 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Again row transformation, R 3 → R 3 − 4 R 1 , ( 1 − 4 1 0 − 18 9 0 18 − 9 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Again by row transformation, R 3 → R 3 + R 2 , ( 1 − 4 1 0 − 18 9 0 0 0 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( x 1 − 4 y 1 + z 1 − 18 y 1 + 9 z 1 0 ) = ( 0 0 0 ) Which is equivalent to the equations, x 1 − 4 y 1 + z 1 = 0 and − 18 y 1 + 9 z 1 = 0 . On solving further it gives, x 1 = 4 y 1 − z 1 and y 1 = 1 2 z 1 which implies, x 1 = 4 2 z 1 − z 1 x 1 = 2 z 1 − z 1 x 1 = z 1 Assume that z 1 = p 1 for some scalar p 1 which implies, x 1 = p 1 y 1 = 1 2 p 1 Therefore, the value of ( x 1 y 1 z 1 ) is, ( x 1 y 1 z 1 ) = ( p 1 1 2 p 1 p 1 ) = p 1 ( 1 1 2 1 ) = 2 p 1 ( 2 1 2 ) Thus, the eigenvector K 1 corresponding to λ 1 = 8 is ( 2 1 2 )

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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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. Substitute the values of

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