Explanation Given: The matrix A = [ 2 − 2 1 1 − 4 1 2 2 − 3 ] . Approach: Let A ne ( n × n ) real matrix. Let v 1 , v 2 , ... , v n be linearly independent eigenvector vector corresponding to eigenvalues λ 1 , λ 2 , ... , λ n . Then, x ( t ) = ∑ k = 1 n C k e λ k t v k is the general solution to the system x ′ = A x . The eigenvector corresponding to the eigenvalue λ is given by the solution to ( A − λ I ) x = 0 . Calculation: Eigenvalues are the roots of characteristic equation of the matrix A = [ 2 − 2 1 1 − 4 1 2 2 − 3 ] . Characteristic equation is r 3 + 5 r 2 − 4 r − 20 = 0 ( r − 2 ) ( r 2 + 7 r + 10 ) = 0 ( r − 2 ) ( r + 2 ) ( r + 5 ) = 0 So, the eigenvalues of the matrix A are 2 , − 2 and − 5 . The eigenvector corresponding to the eigenvalue λ = 2 is given by the solution to ( A − 2 I ) x = 0 [ 0 − 2 1 1 − 6 1 2 2 − 5 ] [ x 1 x 2 x 3 ] = [ 0 0 0 ] The reduced echelon form [ 0 − 2 1 1 − 6 1 2 2 − 5 ] is [ 1 0 − 2 0 1 − 1 2 0 0 0 ] . Then, x 1 − 2 x 3 = 0 , x 2 − x 3 2 = 0 The eigenvector corresponding to the eigenvalue λ = 2 is [ 4 1 2 ]

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Chapter 9.11, Problem 13AP

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The general solution to the linear system x′=Ax for the given matrix A=[2−211−4122−3].

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The general solution to the linear system x ′ = A x for the given matrix A = [ 2 − 2 1 1 − 4 1 2 2 − 3 ] .

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The general solution to the linear system x′=Ax for the given matrix A=[2−211−4122−3].

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