Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.9 Exercises - Page 79: 11

Answer

See solution. Rotating by $90^o$ around origin can be done by multiplying vector $\begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix}$

Work Step by Step

Reflecting points through the $x_1$ then the $x_2$ axis is the same as rotating it around the origin by $180^o$. Lets demonstrate on vector $b=\begin{bmatrix} b_1\\ b_2 \end{bmatrix}$ a) Reflecting through $x_1$ $\begin{bmatrix} 1&0\\ 0&-1 \end{bmatrix}\begin{bmatrix} b_1\\ b_2 \end{bmatrix}=\begin{bmatrix} b_1\\ -b_2 \end{bmatrix}$ Reflecting through $x_2$ $\begin{bmatrix} -1&0\\ 0&1 \end{bmatrix}\begin{bmatrix} b_1\\ -b_2 \end{bmatrix}=\begin{bmatrix} -b_1\\ -b_2 \end{bmatrix}$ Resulting vector is$\begin{bmatrix} -b_1\\ -b_2 \end{bmatrix}$ b) Rotating twice around origin by $90^o$ $\begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix}\begin{bmatrix} b_1\\ b_2 \end{bmatrix}=\begin{bmatrix} -b_2\\ b_1 \end{bmatrix}$ $\begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix}\begin{bmatrix} -b_2\\ b_1 \end{bmatrix}=\begin{bmatrix} -b_1\\ -b_2 \end{bmatrix}$ Resulting vector is$\begin{bmatrix} -b_1\\ -b_2 \end{bmatrix}$, which is the same as part a)
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