Program Explanation Given information: The system is x ′ = [ 1 3 7 0 0 − 1 − 4 0 0 1 3 0 0 − 6 − 14 1 ] x . Explanation: Consider matrix form of the system of equation as A . A = [ 1 3 7 0 0 − 1 − 4 0 0 1 3 0 0 − 6 − 14 1 ] The characteristics equation of the coefficient matrix is, [ 1 − λ 3 7 0 0 − 1 − λ − 4 0 0 1 3 − λ 0 0 − 6 − 14 1 − λ ] = 0 ( λ − 1 ) 4 = 0 λ = 1 , 1 , 1 and 1 The obtain Eigen value has multiplicity 4. The Eigen vector of the coefficient matrix A is in the form as, [ − 1 − λ − 4 0 0 1 3 − λ 0 0 1 2 1 − λ 0 0 1 0 1 − λ ] [ a b c d ] = [ 0 0 0 0 ] The associated Eigen vector with the matrix A is v = [ a b c d ] T . Substitute 1 for λ in coefficient matrix, [ 1 − 1 3 7 0 0 − 1 − 1 − 4 0 0 1 3 − 1 0 0 − 6 − 14 1 − 1 ] [ a b c d ] = [ 0 0 0 0 ] [ 0 3 7 0 0 − 2 − 4 0 0 1 2 0 0 − 6 − 14 0 ] [ a b c d ] = [ 0 0 0 0 ] The scalar equation are shown below, 3 b + 7 c = 0 − 2 b − 4 c = 0 a + 2 b = 0 − 6 a − 14 b = 0 Therefore, the associated Eigen vector is obtained as: u 1 = [ 1 0 0 − 2 ] The other associated Eigen vectors is obtained as, u 2 = [ 1 0 0 0 ] The defect of the Eigen value is 2. The other three associated vectors can be obtained by the given formula, v 2 = ( A − I ) v 3 v 1 = ( A − I ) v 2 The value of ( A − I ) 2 can be obtained as, ( A − I ) 2 = [ 0 3 7 0 0 − 2 − 4 0 0 1 2 0 0 − 6 − 14 0 ] [ 0 3 7 0 0 − 2 − 4 0 0 1 2 0 0 − 6 − 14 0 ] = [ 0 1 2 0 0 0 0 0 0 0 0 0 0 − 2 − 4 0 ] The value of ( A − I ) 3 can be obtained as, ( A − I ) 3 = [ 0 1 2 0 0 0 0 0 0 0 0 0 0 − 2 − 4 0 ] [ 0 3 7 0 0 − 2 − 4 0 0 1 2 0 0 − 6 − 14 0 ] = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ] Therefore, any vector v 3 ≠ 0 is a solution of the equation ( A − I ) 3 v 3 = 0 . Consider the vector v 3 = [ 0 1 0 0 ] The associated vector u = v 3

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Problems on Growing Functions

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Chapter 5.5, Problem 22P

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[13700−1−4001300−6−141]x .

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Chapter 5.5, Problem 21P

Chapter 5.5, Problem 23P

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Chapter 5 Solutions

EBK DIFFERENTIAL EQUATIONS

Ch. 5.1 - Let A=[2347] and B=[3451]. Find (a) 2A+3B; (b)...Ch. 5.1 - Prob. 2P Ch. 5.1 - Find AB and BA given A=[203415] and B=[137032].Ch. 5.1 - Prob. 4P Ch. 5.1 - Prob. 5P Ch. 5.1 - Prob. 6P Ch. 5.1 - Prob. 7P Ch. 5.1 - Prob. 8P Ch. 5.1 - Prob. 9P Ch. 5.1 - Prob. 10P

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Program Description: Purpose of problem is to obtain the general solution of the system x ′ = [ 1 3 7 0 0 − 1 − 4 0 0 1 3 0 0 − 6 − 14 1 ] x .

Solution Summary: The author explains that the problem is to obtain the general solution of the system of equation as A.

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[13700−1−4001300−6−141]x .

Want to see the full answer?

Chapter 5 Solutions