Linear Algebra and Its Applications (5th Edition)

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

Chapter 2 - Matrix Algebra - 2.7 Exercises - Page 146: 8

Answer

$\left[\begin{array}{ccc}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 2 \sqrt{2}+3 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 7-5 \sqrt{2} \\ 0 & 0 & 1\end{array}\right]$

Work Step by Step

\[ \begin{array}{l} {\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right] \rightarrow\left[\begin{array}{lll} 1 & 0 & -3 \\ 0 & 1 & -7 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right]} \\ {\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right] \rightarrow\left[\begin{array}{lll} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & -3 \\ 0 & 1 & -7 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right]} \\ {\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right] \rightarrow\left[\begin{array}{lll} 1 & 0 & 3 \\ 0 & 1 & 7 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & -3 \\ 0 & 1 & -7 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right]} \end{array} \] rotation of points accomplished by first translating the figure by $-P$, then rotate about the origin, then again translate back by P. \[ \left[\begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 7 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -3 \\ 0 & 1 & -7 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 2 \sqrt{2}+3 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 7-5 \sqrt{2} \\ 0 & 0 & 1 \end{array}\right] \]
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