Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 2 - Matrices - 2.1 Operations with Matrices - 2.1 Exercises - Page 50: 70



Work Step by Step

We have \begin{align*} AB=\left[\begin{array}{cc}{\cos \alpha} &{-\sin \alpha} \\{ \sin \alpha}& {\cos \alpha} \end{array}\right]\left[\begin{array}{cc}{\cos \beta} &{-\sin \beta} \\{ \sin \beta}& {\cos \beta} \end{array}\right]\\ = \left[\begin{array}{cc}{\cos \alpha\cos \beta-\sin \alpha\sin \beta} &{-\cos \alpha\sin\beta-\sin \alpha\cos \beta} \\{ \sin\alpha\cos \beta+\cos \alpha\sin \beta}& {-\sin \alpha\sin\beta+\cos\alpha\cos\beta} \end{array}\right] \end{align*} and \begin{align*} BA= \left[\begin{array}{cc}{\cos \alpha} &{-\sin \alpha} \\{ \sin \alpha}& {\cos \alpha} \end{array}\right]\left[\begin{array}{cc}{\cos \beta} &{-\sin \beta} \\{ \sin \beta}& {\cos \beta} \end{array}\right]\\ = \left[\begin{array}{cc}{\cos \alpha\cos \beta-\sin \alpha\sin \beta} &{-\cos \alpha\sin\beta-\sin \alpha\cos \beta} \\{ \sin\alpha\cos \beta+\cos \alpha\sin \beta}& {-\sin \alpha\sin\beta+\cos\alpha\cos\beta} \end{array}\right] \end{align*}. Which shows that $AB=BA$.
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