Program Explanation Given information: The system is x ′ = [ − 2 − 9 0 1 4 0 1 3 1 ] x . Explanation: Consider matrix form of the system of equation as A . A = [ − 2 − 9 0 1 4 0 1 3 1 ] The characteristics equation of the coefficient matrix is, [ − 2 − λ − 9 0 1 4 − λ 0 1 3 1 − λ ] = 0 ( − 2 − λ ) [ 4 − λ 0 3 1 − λ ] − ( − 9 ) [ 1 0 1 1 − λ ] + 0 = 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, [ − 2 − λ − 9 0 1 4 − λ 0 1 3 1 − λ ] [ a b c ] = [ 0 0 0 ] The associated Eigen vector with the matrix Ais v = [ a b c ] T . Substitute 1 for λ in coefficient matrix, [ − 2 − 1 − 9 0 1 4 − 1 0 1 3 1 − 1 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] [ − 3 − 9 0 1 3 0 1 3 0 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] The scalar equation are shown below, − 3 a 1 − 9 b 1 = 0 a 1 + 3 b 1 = 0 a 1 + 3 b 1 = 0 If c 1 is independent the associated Eigen vector is shown below: u 1 = [ 3 − 1 0 ] Similarly the second case the associated Eigen vector is shown below: u 2 = [ 0 0 1 ] The one Eigen vector is missing, therefore the defect is 1. The other two associated vectors can be obtained by the given formula, v 1 = ( A − I ) v 2 The value of ( A − I ) 2 can be obtained as, ( A − I ) 2 = [ − 3 − 9 0 1 3 0 1 3 0 ] [ − 3 − 9 0 1 3 0 1 3 0 ] = [ 0 0 0 0 0 0 0 0 0 ] Therefore, any vector v ≠ 0 is a solution of the equation ( A − I ) 3 v = 0

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

5th Edition

ISBN: 9780134872971

Author: Edwards, C., Penney, David, Calvis

Publisher: PEARSON

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Chapter 5.5, Problem 15P

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[−2−90140131]x .

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Chapter 5.5, Problem 14P

Chapter 5.5, Problem 16P

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Chapter 5 Solutions

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

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 ′ = [ − 2 − 9 0 1 4 0 1 3 1 ] x .

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[−2−90140131]x .

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Chapter 5 Solutions