Program Explanation Given information: The system is x ′ = [ − 3 0 − 4 − 1 − 1 − 1 1 0 1 ] x . Explanation: Consider matrix form of the system of equation as A . A = [ − 3 0 − 4 − 1 − 1 − 1 1 0 1 ] The characteristics equation of the coefficient matrix is, [ − 3 − λ 0 − 4 − 1 − 1 − λ − 1 1 0 1 − λ ] = 0 ( − 3 − λ ) [ − 1 − λ − 1 0 1 − λ ] − 0 + ( − 4 ) [ − 1 − 1 − λ 1 0 ] = 0 − λ 3 − 3 λ 2 + λ + 3 − 4 − 4 λ = 0 λ 3 + 3 λ 2 + 3 λ+1 = 0 Further solve the above equation as shown below. λ 3 + 3 λ 2 + 3 λ+1 = 0 ( λ + 1 ) 3 = 0 λ = − 1 , − 1 , and − 1 The Eigen vector of the coefficient matrix A is in the form as, [ − 3 − λ 0 − 4 − 1 − 1 − λ − 1 1 0 1 − λ ] [ a b c ] = [ 0 0 0 ] The associated Eigen vector with the matrix A is v = [ a b c ] T . Substitute − 1 for λ in coefficient matrix, [ − 3 + 1 0 − 4 − 1 − 1 + 1 − 1 1 0 1 + 1 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] [ − 2 0 − 4 − 1 0 − 1 1 0 2 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] The scalar equation are shown below, − 2 a 1 − 4 c 1 = 0 − a 1 − c 1 = 0 a 1 + 2 c 1 = 0 Therefore, the associated Eigen vector is shown below. v 1 = [ 0 1 0 ] The two Eigen vector is missing, therefore the defect is 2. The other two associated vectors can be obtained by the given formula, v 2 = ( A − λ I ) 2 v 3 v 1 = ( A − λ I ) 2 v 2 Consider the vector v 3 = [ 1 0 0 ] Now associated vector v 2 can be obtained as, v 2 = [ − 3 − λ 0 − 4 − 1 − 1 − λ − 1 1 0 1 − λ ] 2 [

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 11P

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Program Description:Purpose of problem is to find general solution of the system x′=[−30−4−1−1−1101]x .

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Chapter 5.5, Problem 10P

Chapter 5.5, Problem 12P

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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 find general solution of the system x ′ = [ − 3 0 − 4 − 1 − 1 − 1 1 0 1 ] x .

Solution Summary: The author explains that the problem is to find general solution of the system of equation as shown below.

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Program Description:Purpose of problem is to find general solution of the system x′=[−30−4−1−1−1101]x .

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Chapter 5 Solutions