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: 7

Answer

$A=\begin{bmatrix} -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \end{bmatrix}$

Work Step by Step

$A=\begin{bmatrix} T(e_1) & ... & T(e_n)\\ \end{bmatrix}$ $e_1=\begin{bmatrix} 1\\ 0 \end{bmatrix}$ and $e_2=\begin{bmatrix} 0\\ 1 \end{bmatrix}$ $T(e_1)=\begin{bmatrix} -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{bmatrix}$ and $T(e_2)=\begin{bmatrix} 1&0\\ 0&-1 \end{bmatrix}\begin{bmatrix} -\sin(-\frac{3\pi}{4})\\ \cos(-\frac{3\pi}{4}) \end{bmatrix}=\begin{bmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{bmatrix}$ $A=\begin{bmatrix} -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \end{bmatrix}$
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