Explanation The given matrix A is [ 2 − 1 0 5 2 4 0 1 2 ] It is known that if A be an n × n matrix then A number λ is said to be an eigenvalue of A if there exists a nonzero solution vector K of the linear system. A K = λ K The solution vector K is said to be an eigenvector corresponding to the Eigenvalue λ . Eigenvalues and eigenvectors are also called characteristic values and characteristic vectors, respectively. Therefore, the eigenvalues of A are the roots of characteristic equation det ( A − λ I ) = 0 . det [ A − λ I ] = det [ [ 2 − 1 0 5 2 4 0 1 2 ] − λ [ 1 0 0 0 1 0 0 0 1 ] ] = [ 2 − λ − 1 0 5 2 − λ 4 0 1 2 − λ ] = ( 2 − λ ) det [ 2 − λ 4 1 2 − λ ] − ( − 1 ) det [ 5 4 0 2 − λ ] + 0 det [ 5 2 − λ 0 1 ] Further solve the above expression. det [ A − λ I ] = ( 2 − λ ) ( λ 2 − 4 λ ) − ( − 1 ) ⋅ 5 ( − λ + 2 ) + 0 ⋅ 5 = − λ 3 + 6 λ 2 − 13 λ + 10 Now solve the above cubic equation as follows to obtain the Eigen values. − ( λ − 2 ) ( λ 2 − 4 λ + 5 ) = 0 − ( λ − 2 ) = 0 or ( λ 2 − 4 λ + 5 ) = 0 λ =2 or ( λ 2 − 4 λ + 5 ) = 0 Solve ( λ 2 − 4 λ + 5 ) = 0 by completing the square. ( λ 2 − 4 λ + 5 ) = 0 λ 2 − 4 λ + 4 = − 1 ( λ − 2 ) 2 = − 1 or ( λ − 2 ) = − 1 = ± i λ = i + 2 or λ = − i + 2 So the Eigen values of the given matrix are complex having value ( 2, 2 + i , 2 − i ) . Now for λ = 2 Eigen vector is calculated as follows: [ A − λ I ] = [ 2 − 1 0 5 2 4 0 1 2 ] − 2 [ 1 0 0 0 1 0 0 0 1 ] = [ 2 − 1 0 5 2 4 0 1 2 ] − [ 2 0 0 0 2 0 0 0 2 ] = [ 0 − 1 0 5 0 4 0 1 0 ] Reduce the obtained matrix in row echelon form as shown below. [ A − λ I ] = [ 0 − 1 0 5 0 4 0 1 0 ] Apply the transformation R 1 ↔ R 2 in matrix [ A − λ I ] = [ 5 0 4 0 − 1 0 0 1 0 ] Apply the transformation R 3 ← R 3 + 1 ⋅ R 2 in the above matrix. [ A − λ I ] = [ 5 0 4 0 − 1 0 0 0 0 ] Apply the transformation R 2 ← − 1 ⋅ R 2 in the above matrix [ A − λ I ] = [ 5 0 4 0 1 0 0 0 0 ] Apply the transformation R 1 ← 1 5 ⋅ R 1 in the above matrix [ A − λ I ] = [ 1 0 4 5 0 1 0 0 0 0 ] Now the corresponding equation can be written as follows: [ A − 2 I ] [ x y z ] = [ 0 0 0 ] [ 1 0 4 5 0 1 0 0 0 0 ] [ x y z ] = [ 0 0 0 ] The equation obtained from the above expression is as follows: [ x + 4 5 z = 0 y = 0 ] or [ y = 0 x = − 4 5 z ] Now the eigen vector can be written as follows: [ − 4 5 z 0 z ] Substitute z = 1 in the above matrix. [ − 4 5 0 1 ] Eigen vector for λ = 2 is obtained as follows: [ − 4 5 0 1 ] Now for λ = 2 + i Eigen vector is given as follows: [ A − ( 2 + i ) I ] = [ 2 − 1 0 5 2 4 0 1 2 ] − ( 2 + i ) [ 1 0 0 0 1 0 0 0 1 ] = [ 2 − 1 0 5 2 4 0 1 2 ] − [ 2 + i 0 0 0 2 + i 0 0 0 2 + i ] = [ − i − 1 0 5 − i 4 0 1 − i ] Reduce the obtained matrix in row echelon form as follows: [ A − ( 2 + i ) I ] = [ − i − 1 0 5 − i 4 0 1 − i ] Apply the transformation R 2 ← R 2 − 5 i ⋅ R 1 in the above matrix

Differential Equations with Boundary-Value Problems

9th Edition

ISBN: 9781337632515

Author: Dennis G. Zill

Publisher: Cengage Learning US

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Chapter B, Problem 56E

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The verification that the matrix [2−10524012] has complex eigenvalues and find the eigenvectors of the matrix.

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Chapter B, Problem 55E

Chapter B, Problem 57E

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Chapter B Solutions

Differential Equations with Boundary-Value Problems

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The verification that the matrix [ 2 − 1 0 5 2 4 0 1 2 ] has complex eigenvalues and find the eigenvectors of the matrix.

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The verification that the matrix [2−10524012] has complex eigenvalues and find the eigenvectors of the matrix.

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Chapter B Solutions