Explanation Given: The given system is, d y d t = − 6 x + 5 y d y d t = − 5 x + 4 y Calculation: For the given system, the coefficient matrix is A = ( − 6 5 − 5 4 ) The characteristic equation is | ( A − λ I ) | = 0 where I is the identity matrix. Substitute the values of matrix A and I in above expression. | − 6 − λ 5 − 5 4 − λ | = 0 ( − 6 − λ ) ( 4 − λ ) + 25 = 0 1 + 2 λ + λ 2 = 0 ( λ + 1 ) 2 = 0 Therefore, the eigenvalues of the above system is λ = − 1 with multiplicity 2 . Obtain the eigenvector K 1 corresponding to the eigenvalue λ = − 1 . Solve the equation ( A − λ I ) K 1 = 0 , where K 1 is given by ( x 1 y 1 ) . Substitute the values of A , λ and K 1 in the above equation gives, [ ( − 6 5 − 5 4 ) − ( − 1 0 0 − 1 ) ] ⋅ ( x 1 y 1 ) = ( 0 0 ) ( − 5 5 − 5 5 ) ⋅ ( x 1 y 1 ) = ( 0 0 ) ( − 5 x 1 + 5 y 1 − 5 x 1 + 5 y 1 ) = ( 0 0 ) By solving the above equations gives y 1 = x 1 . Assume that x 1 = p 1 for some scalar p 1 , which implies y 1 = p 1 . Thus, the value of ( x 1 y 1 ) is, ( x 1 y 1 ) = ( p 1 p 1 ) = p 1 ( 1 1 ) Therefore, the eigenvector K 1 corresponding to λ = − 1 is ( 1 1 ) . Obtain the second eigenvector K 2 . Solve the equation ( A − λ I ) K 2 = K 1 , where K 2 is given by ( x 2 y 2 )

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 10.2, Problem 22E

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Chapter 10.2, Problem 21E

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