Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.3 - Matrix Operations and Their Applications - Exercise Set - Page 918: 55

Answer

The translated matrix is $B=\left[ \begin{matrix} 0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\ 1 & 1 & \frac{3}{2} & \frac{3}{2} & \frac{7}{2} & \frac{7}{2} \\ \end{matrix} \right]$ The resulting graph is shown below:
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Work Step by Step

To reduce L to half its perimeter, we will multiply each $x\text{-coordinate}$ and $y\text{-coordinate}$ by $\frac{1}{2}$. That is, $\begin{align} & B=\frac{1}{2}\times \left[ \begin{matrix} 0 & 3 & 3 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & \frac{5}{2} & \frac{5}{2} \\ \end{matrix} \right] \end{align}$ Now, we will add 1 to each $y\text{-coordinate}$ to move the reduced image 1 unit up. That is, $\begin{align} & B=\left[ \begin{matrix} 0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & \frac{5}{2} & \frac{5}{2} \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\ 1 & 1 & \frac{3}{2} & \frac{3}{2} & \frac{7}{2} & \frac{7}{2} \\ \end{matrix} \right] \end{align}$
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