Chapter 2.4, Problem 20E

To determine

To find: the coordinate of the triangle using matrices and graph the pre image and image on same coordinate.

Answer to Problem 20E

  $D^{\prime }\left(-2,4\right),E^{\prime }\left(-6,2\right),F^{\prime }\left(-3,-4\right)\text{ and }{G}^{\prime }\left(1,-2\right)$ .

Explanation of Solution

Given:

The rectangle with vertices:

  $D\left(2,4\right),E\left(6,2\right),F\left(3,-4\right)\text{ and }G\left(-1,-2\right)$ .

Concept used:

The image of linear transformation or matrix is the span of the vector of the linear transformation.

It can be written as $Im\left(A\right)$ .

For example, consider the matrix $A$ which is equal to:

  $\left[\begin{array}{cc}1& 0\\ 0& 2\\ 0& 1\end{array}\right]$

Multiplying this by $2\times 1$ gives a $3\times 1$ matrix. However, regardless of what vector is chosen to multiply by, there are some vectors that can’t be the result.

Result is not image of $A$ . The vector that are possible belong to the span of $A$ . In this case, the span be represented by a parametrized matrix, where $t\text{ and }s$ can be any number:

  $\left[\begin{array}{cc}1& 0\\ 0& 2\\ 0& 1\end{array}\right]\times \left[\begin{array}{c}s\\ t\end{array}\right]=\left[\begin{array}{c}s\\ 2t\\ t\end{array}\right]=Im\left(A\right)$ .

Matrices multiplication:

  $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]=\left[\begin{array}{cc}ae+ag& af+bh\\ ce+dg& cf+dh\end{array}\right].$

Calculation:

The rectangle with vertices:

  $D\left(2,4\right),E\left(6,2\right),F\left(3,-4\right)\text{ and }G\left(-1,-2\right)$ .

First write the vertex matrix as:

  $\left[\begin{array}{cccc}\begin{array}{l}D\\ 2\end{array}& \begin{array}{l}E\\ 6\end{array}& \begin{array}{l}F\\ 3\end{array}& \begin{array}{l}G\\ -1\end{array}\\ 4& 2& -4& -2\end{array}\right]$ .

Reflected over the $y$ -axis and then $R_{y-axis}$ is:

  $\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]$ .

Multiply by the $y$ -axis reflection matrix.

  $$

  $\begin{array}{l}\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cccc}2& 6& 3& -1\\ 4& 2& -4& -2\end{array}\right]\\ \Rightarrow \left[\begin{array}{cccc}-2& -6& -3& 1\\ 4& 2& -4& -2\end{array}\right].\end{array}$

Graph of the two rectangles is:

Hence, the vertices of the image are $D^{\prime }\left(-2,4\right),E^{\prime }\left(-6,2\right),F^{\prime }\left(-3,-4\right)\text{ and }{G}^{\prime }\left(1,-2\right)$ .