Chapter 8.8, Problem 15E

To determine

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

Answer to Problem 15E

The resultant matrix reflects over all three 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 sign of $x\text{-, }y\text{- and }z\text{-}$ coordinates are changed after transformation. Thus the resultant matrix reflects over all three planes.