Program Explanation Given information: The matrix A is shown below. A = [ 3 0 − 3 5 0 7 3 0 − 3 ] Explanation: Obtain the square of matrix A as shown below. A 2 = A ⋅ A = [ 3 0 − 3 5 0 7 3 0 − 3 ] [ 3 0 − 3 5 0 7 3 0 − 3 ] = [ 0 0 0 36 0 − 36 0 0 0 ] Obtain the cube of matrix A as shown below. A 3 = A ⋅ A 2 = [ 0 0 0 36 0 − 36 0 0 0 ] [ 3 0 − 3 5 0 7 3 0 − 3 ] = [ 0 0 0 0 0 0 0 0 0 ] As cube of matrix A is zero matrix, therefore matrix A is nilpotent A n = 0 for n ≥ 3 . So, evaluate the exponential matrix e A t as shown below. e A t = I + A t + A 2 t 2 2 ! + A 3 t 3 3 ! + ⋯ + A n t n n ! + ⋯ = I + A t + A 2 t 2 2 ! + A 3 t 3 3 ! + A 3 A t 4 4 !

5th Edition

ISBN: 9780135988510

Author: Edwards

Publisher: PEARSON CO

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Question

Chapter 5.6, Problem 24P

Program Plan Intro

Program Description: Purpose of the problem is to obtain value of an exponential matrix eAt by the use of nilpotent property of the matrix A .

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Chapter 5.6, Problem 23P

Chapter 5.6, Problem 25P

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Chapter 5 Solutions

EP DIFF.EQUAT.+BOUND.VALUE,...UPD.-ACC.

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 the problem is to obtain value of an exponential matrix e A t by the use of nilpotent property of the matrix A .

Solution Summary: The author explains that the problem is to obtain value of an exponential matrix eAt by the use of nilpotent property.

Program Description: Purpose of the problem is to obtain value of an exponential matrix eAt by the use of nilpotent property of the matrix A .

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Chapter 5 Solutions