Program Explanation Given information: Matrices A and B are square matrices of order n × n such that matrices commute with each other. A B = B A The equation e A + B = e A e B is true for the commuting matrices A and B . Explanation: Evaluate the exponential of null square matrix as, e 0 = I + 0 t + 0 2 t 2 2 ! + ⋯ + 0 n t n n ! + ⋯ = I Therefore, exponential of null square matrix is an identity matrix

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

5th Edition

ISBN: 9780321816252

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

Publisher: PEARSON

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Question

Chapter 5.6, Problem 32P

Program Plan Intro

Program Description: Purpose of the problem is to prove that (eA)−1=e−A and matrix eA is nonsingular for every square matrix A .

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Chapter 5.6, Problem 31P

Chapter 5.6, Problem 33P

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Chapter 5 Solutions

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

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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Knowledge Booster

Program Description: Purpose of the problem is to prove that ( e A ) − 1 = e − A and matrix e A is nonsingular for every square matrix A .

Solution Summary: The author explains that the exponential of null square matrix is an identity matrix, and the matrix e-A is the inverse of it.

Program Description: Purpose of the problem is to prove that (eA)−1=e−A and matrix eA is nonsingular for every square matrix A .

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Chapter 5 Solutions