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: 7

Answer

$\left[\begin{array}{ccc}\frac{1}{2} & -\frac{\sqrt{3}}{2} & 3+4 \sqrt{3} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & -3 \sqrt{3}+4 \\ 0 & 0 & 1\end{array}\right]$

Work Step by Step

\[ \begin{array}{l} {\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right] \rightarrow\left[\begin{array}{cccc} 1 & 0 & -6 \\ 0 & 1 & -8 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right]} \\ {\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right] \rightarrow\left[\begin{array}{cccc} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -6 \\ 0 & 1 & -8 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ y \\ 1 \end{array}\right]} \\ {\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right] \rightarrow\left[\begin{array}{lll} 1 & 0 & 6 \\ 0 & 1 & 8 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -6 \\ 0 & 1 & -8 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right]} \end{array} \] rotation of points accomplished by first translating the figure by $-P$, then rotate about the origin, then again translate back by P. \[ \left[\begin{array}{ccc} 1 & 0 & 6 \\ 0 & 1 & 8 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -6 \\ 0 & 1 & -8 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 3+4 \sqrt{3} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & -3 \sqrt{3}+4 \\ 0 & 0 & 1 \end{array}\right] \] Required matrix for composite transformation.
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