Explanation Given: The given matrix is: A = [ − 1 − 8 1 − 1 − 3 2 − 4 − 16 7 ] Formula used: Compute the characteristics polynomial p ( r ) = | A − r I | . Find the zeroes of p ( r ) and express it as p ( r ) = ( r 1 − r ) m 1 ( r 2 − r ) m 2 ⋯ ( r k − r ) m k here r 1 , r 2 , ⋯ r k are the eigenvalues and m 1 , m 2 ⋯ , m k are their multiplicities. For each eigenvalue r i find m i linearly independent generalized eigenvectors by applying the Gauss-Jordan algorithm to the system ( A − r i I ) m i u = 0 . Then, assemble the fundamental matrix X ( t ) and find the general solution: x ( t ) = X 1 ( t ) + X 2 ( t ) + X 3 ( t ) Calculation: Consider the given matrix: A = [ − 1 − 8 1 − 1 − 3 2 − 4 − 16 7 ] The characteristics polynomial is: | [ − 1 − 8 1 − 1 − 3 2 − 4 − 16 7 ] − r [ 1 0 0 0 1 0 0 0 1 ] | = 0 | [ − 1 − 8 1 − 1 − 3 2 − 4 − 16 7 ] − [ r 0 0 0 r 0 0 0 r ] | = 0 | [ − 1 − r − 8 1 − 1 − 3 − r 2 − 4 − 16 7 − r ] | = 0 ( − 1 − r ) ⋅ [ − 3 − r 2 − 16 7 − r ] + 8 ⋅ [ − 1 2 − 4 7 − r ] + 1 ⋅ [ − 1 − 3 − r − 4 − 16 ] = 0 Further simplify as: ( − 1 − r ) [ ( − 3 − r ) ( 7 − r ) + 32 ] + 8 [ − ( 7 − r ) + 8 ] + [ 16 + 4 ( − 3 − r ) ] = 0 ( r − 1 ) 3 = 0 r = 1 , 1 , 1 The eigenvalues is 1 with multiplicity 3. The eigenvector corresponding to the eigenvalue r = 1 : ( [ − 1 − 8 1 − 1 − 3 2 − 4 − 16 7 ] − 1 [ 1 0 0 0 1 0 0 0 1 ] ) [ u 1 u 2 u 3 ] = [ 0 0 0 ] ( [ − 1 − 8 1 − 1 − 3 2 − 4 − 16 7 ] − [ 1 0 0 0 1 0 0 0 1 ] ) [ u 1 u 2 u 3 ] = [ 0 0 0 ] [ − 2 − 8 1 − 1 − 4 2 − 4 − 16 6 ] [ u 1 u 2 u 3 ] = [ 0 0 0 ] It reduce using row-echelon, we get: [ − 2 − 8 1 − 1 − 4 2 − 4 − 16 6 ] → [ 1 4 0 0 0 1 0 0 0 ] It indicates as, u 1 + 4 u 2 = 0 and u 3 = 0

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Chapter 9.8, Problem 20E

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The general solution for the given matrix.