Explanation Consider the given matrix: A = [ 1 3 − 2 0 7 − 4 0 9 − 5 ] Substitute the value of matrix A in the characteristic equation and solve for r : | [ 1 3 − 2 0 7 − 4 0 9 − 5 ] − r [ 1 0 0 0 1 0 0 0 1 ] | = 0 | [ 1 3 − 2 0 7 − 4 0 9 − 5 ] − [ r 0 0 0 r 0 0 0 r ] | = 0 | [ 1 − r 3 − 2 0 7 − r − 4 0 9 − 5 − r ] | = 0 ( 1 − r ) ⋅ | 7 − r − 4 9 − 5 − r | = 0 Further simplify as: ( 1 − r ) ⋅ [ ( 7 − r ) ( − 5 − r ) + 36 ] = 0 − ( 1 − r ) [ ( 7 − r ) ( 5 + r ) − 36 ] = 0 − ( 1 − r ) [ 35 + 2 r − r 2 − 36 ] = 0 − ( 1 − r ) [ r 2 − 2 r + 1 ] = 0 Further simplify as, ( r − 1 ) [ r 2 − 2 r + 1 ] = 0 ( r − 1 ) ( r − 1 ) 2 = 0 ( r − 1 ) 3 = 0 The factor is: ( r − 1 ) 3 = 0 r = 1 The eigenvalues are r 1 , 2 , 3 = 1 . Therefore, the eigenvalues are repeated. Find the eigenvector corresponding to eigenvalue r 1 = 1 by using the formula ( A − r I ) u = 0 : ( A − r I ) u = ( [ 1 3 − 2 0 7 − 4 0 9 − 5 ] − 1 [ 1 0 0 0 1 0 0 0 1 ] ) [ u 1 u 2 u 3 ] = ( [ 1 3 − 2 0 7 − 4 0 9 − 5 ] − [ 1 0 0 0 1 0 0 0 1 ] ) [ u 1 u 2 u 3 ] = [ 0 3 − 2 0 6 − 4 0 9 − 6 ] [ u 1 u 2 u 3 ] The matrix equation is, [ 0 3 − 2 0 6 − 4 0 9 − 6 ] [ u 1 u 2 u 3 ] = [ 0 0 0 ] It can be write in scalar equation as, 3 u 2 − 2 u 3 = 0 and u 1 is free

Fundamentals of Differential Equations and Boundary Value Problems

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ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 9.5, Problem 40E

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The general solution for the system x′(t)=[13−207−409−5]x(t).

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Chapter 9.5, Problem 39E

Chapter 9.5, Problem 41E

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The general solution for the system x ′ ( t ) = [ 1 3 − 2 0 7 − 4 0 9 − 5 ] x ( t ) .

Solution Summary: The author explains that the general solution for the given system is xprime (t)=c_1et

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To find:

The general solution for the system x′(t)=[13−207−409−5]x(t).

Want to see the full answer?

Chapter 9 Solutions

The general solution for the system x ′ ( t ) = [ 1 3 − 2 0 7 − 4 0 9 − 5 ] x ( t ) .

Solution Summary: The author explains that the general solution for the given system is xprime (t)=c_1et

Videos

To find: The general solution for the system x′(t)=[13−207−409−5]x(t).

Want to see the full answer?

Chapter 9 Solutions

To find:

The general solution for the system x′(t)=[13−207−409−5]x(t).