Chapter 2.4, Problem 15E

To determine

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

Answer to Problem 15E

  $\left[\begin{array}{ccc}\begin{array}{l}{W}^{\prime }\\ 1\end{array}& \begin{array}{l}{X}^{\prime }\\ 4\end{array}& \begin{array}{l}{Y}^{\prime }\\ 6\end{array}\\ -2& 3& -3\end{array}\right]$ .

Explanation of Solution

Given:

Triangle $WXY$ with vertex matrix $\left[\begin{array}{ccc}-2& 1& 3\\ 0& 5& -1\end{array}\right]$ .

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:

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

Addition of matrix:

  $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]+\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]=\left[\begin{array}{cc}a+e& b+f\\ c+g& d+h\end{array}\right]$ .

Calculation:

Triangle $WXY$ with vertex matrix $\left[\begin{array}{ccc}-2& 1& 3\\ 0& 5& -1\end{array}\right]$ .

  $\begin{array}{c}x-\text{coordinate}\\ y-\text{coordinate}\end{array}\left[\begin{array}{ccc}\begin{array}{l}W\\ -2\end{array}& \begin{array}{l}X\\ 1\end{array}& \begin{array}{l}Y\\ 3\end{array}\\ 0& 5& -1\end{array}\right]$ .

Translated $3$ units right and $2$ units down then the translation matrix is:

  $\left[\begin{array}{ccc}3& 3& 3\\ -2& -2& -2\end{array}\right]$ .

Add the two matrices:

  $\begin{array}{l}\left[\begin{array}{ccc}-2& 1& 3\\ 0& 5& -1\end{array}\right]+\left[\begin{array}{ccc}3& 3& 3\\ -2& -2& -2\end{array}\right]\\ \Rightarrow \left[\begin{array}{ccc}\begin{array}{l}{W}^{\prime }\\ 1\end{array}& \begin{array}{l}{X}^{\prime }\\ 4\end{array}& \begin{array}{l}{Y}^{\prime }\\ 6\end{array}\\ -2& 3& -3\end{array}\right].\end{array}$

Graph the points represented by the resulting matrix.

Hence, $\left[\begin{array}{ccc}\begin{array}{l}{W}^{\prime }\\ 1\end{array}& \begin{array}{l}{X}^{\prime }\\ 4\end{array}& \begin{array}{l}{Y}^{\prime }\\ 6\end{array}\\ -2& 3& -3\end{array}\right]$ .