Chapter 8.8, Problem 14E

To determine

To describe: the results obtained after transforming the figure using the given matrix, and also graph each image.

Answer to Problem 14E

After transformation the $x$ coordinates of each point remains same, while the sign of $y\text{- and }z\text{-}$ coordinates are changed. Thus the reflection is seen on $xy\text{ and }xz$ planes.

Explanation of Solution

Given information:

The figure representing the prism is $\left(\text{ let }P\right)$ is shown below.

  

The matrix used for transformation $\left(\text{ let }T\right)$ is shown below.

  $\left[\begin{array}{ccc}\begin{array}{l}1\\ 0\\ 0\end{array}& \begin{array}{l}0\\ -1\\ 0\end{array}& \begin{array}{l}0\\ 0\\ -1\end{array}\end{array}\right]$

Calculation:

The matrix of prism $P$ is shown below.

  $P=\left[\begin{array}{rrrrrrrr}\hfill \begin{array}{l}0\\ 0\\ 1\end{array}& \hfill \begin{array}{l}0\\ 3\\ 2\end{array}& \hfill \begin{array}{l}0\\ 5\\ 5\end{array}& \hfill \begin{array}{l}0\\ 0\\ 4\end{array}& \hfill \begin{array}{l}2\\ 0\\ 1\end{array}& \hfill \begin{array}{l}2\\ 0\\ 4\end{array}& \hfill \begin{array}{l}2\\ 3\\ 5\end{array}& \hfill \begin{array}{l}2\\ 3\\ 2\end{array}\end{array}\right]$

The matrix of prism is translated by adding $P\text{ and }T$ . Thus the translated matrix of the prism $\left(\text{let }T\text{'}\right)$ is as follows:

  $\begin{array}{c}T\text{'}=\left[\begin{array}{rrrrrrrr}\hfill \begin{array}{l}0\\ 0\\ 1\end{array}& \hfill \begin{array}{l}0\\ 3\\ 2\end{array}& \hfill \begin{array}{l}0\\ 5\\ 5\end{array}& \hfill \begin{array}{l}0\\ 0\\ 4\end{array}& \hfill \begin{array}{l}2\\ 0\\ 1\end{array}& \hfill \begin{array}{l}2\\ 0\\ 4\end{array}& \hfill \begin{array}{l}2\\ 3\\ 5\end{array}& \hfill \begin{array}{l}2\\ 3\\ 2\end{array}\end{array}\right]\cdot \left[\begin{array}{ccc}\begin{array}{l}1\\ 0\\ 0\end{array}& \begin{array}{l}0\\ -1\\ 0\end{array}& \begin{array}{l}0\\ 0\\ -1\end{array}\end{array}\right]\\ =\left[\begin{array}{rrrrrrrr}\hfill \begin{array}{l}0\\ 0\\ -1\end{array}& \hfill \begin{array}{l}0\\ -3\\ -2\end{array}& \hfill \begin{array}{l}0\\ -5\\ -5\end{array}& \hfill \begin{array}{l}0\\ 0\\ -4\end{array}& \hfill \begin{array}{l}2\\ 0\\ -1\end{array}& \hfill \begin{array}{l}2\\ 0\\ -4\end{array}& \hfill \begin{array}{l}2\\ -3\\ -5\end{array}& \hfill \begin{array}{l}2\\ -3\\ -2\end{array}\end{array}\right]\end{array}$

The graph of the translated matrix is as follows:

  

This shows that $x$ coordinates of each point remains same even after transformation while the sign of $y\text{- and }z\text{-}$ coordinates are changed. Thus the reflection is seen on $xy\text{ and }xz$ planes.