Chapter 2.4, Problem 17E

To determine

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

Answer to Problem 17E

The vertices of the image are $C^{\prime }\left(0,5\right),D^{\prime }\left(4,4\right),E^{\prime }\left(8,5\right)\text{ and }{F}^{\prime }\left(4,1\right)$ ..

Explanation of Solution

Given:

Square with vertices:

  $C\left(-3,1\right),D\left(1,5\right),E\left(5,1\right)\text{ and }F\left(1,-3\right)$ .

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:

Square with vertices:

  $C\left(-3,1\right),D\left(1,5\right),E\left(5,1\right)\text{ and }F\left(1,-3\right)$ .

First write the square as a vertex matrix.

  $\begin{array}{c}x-\text{coordinate}\\ y-\text{coordinate}\end{array}\left[\begin{array}{cccc}\begin{array}{l}C\\ -3\end{array}& \begin{array}{l}D\\ 1\end{array}& \begin{array}{l}E\\ 5\end{array}& \begin{array}{l}F\\ 1\end{array}\\ 1& 5& 1& -3\end{array}\right]$ .

Translated $3$ units right and $4$ unit up.

Translated matrix is:

  $\left[\begin{array}{cccc}3& 3& 3& 3\\ 4& 4& 4& 4\end{array}\right]$ .

Add the two matrices:

  $\begin{array}{l}\left[\begin{array}{cccc}-3& 1& 5& 1\\ 1& 5& 1& -3\end{array}\right]+\left[\begin{array}{cccc}3& 3& 3& 3\\ 4& 4& 4& 4\end{array}\right].\\ \Rightarrow \left[\begin{array}{cccc}\begin{array}{l}{C}^{\prime }\\ 0\end{array}& \begin{array}{l}{D}^{\prime }\\ 4\end{array}& \begin{array}{l}{E}^{\prime }\\ 8\end{array}& \begin{array}{l}{F}^{\prime }\\ 4\end{array}\\ 5& 4& 5& 1\end{array}\right].\end{array}$

Graph of the translated figure:

Hence, the vertices of the image are $C^{\prime }\left(0,5\right),D^{\prime }\left(4,4\right),E^{\prime }\left(8,5\right)\text{ and }{F}^{\prime }\left(4,1\right)$ .