Chapter 2.4, Problem 6CFU

To determine

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

Answer to Problem 6CFU

  $\left[\begin{array}{cccc}\begin{array}{l}{A}^{\prime }\\ -2\end{array}& \begin{array}{l}{B}^{\prime }\\ 2\end{array}& \begin{array}{l}{C}^{\prime }\\ 2\end{array}& \begin{array}{l}{D}^{\prime }\\ -2\end{array}\\ 1& 1& -3& -3\end{array}\right]$ .

Explanation of Solution

Given:

  $A\left(-1,3\right),B\left(3,3\right),C\left(3,-1\right)\text{ and }D\left(-1,-1\right)$ .

Concept used:

Scale factor:

In the equation like $y=Cx$ scalar factor is $C$ .

Which is multiplies the number.

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 $ABCD$ has vertices

  $\begin{array}{l}\left[\begin{array}{ccc}-2& 1& 0\\ 5& 3& -2\end{array}\right]\\ \Rightarrow 1.5\left[\begin{array}{ccc}-2& 1& 0\\ 5& 3& -2\end{array}\right].\\ \Rightarrow \left[\begin{array}{ccc}-3& 1.5& 0\\ 7.5& 4.5& -3\end{array}\right].\end{array}$

The vertices of the image are $J^{\prime }\left(-3,7.5\right),K^{\prime }\left(1.5,4.5\right)and{L}^{\prime }\left(0,-3\right)$ .

$A\left(-1,3\right),B\left(3,3\right),C\left(3,-1\right)\text{ and }D\left(-1,-1\right)$ the matrix representing the vertices square $ABCD$ will be $2\times 4$ matrix.

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

It translates $1$ unit left and $2$ units down.

The translation matrix is:

  $\left[\begin{array}{cccc}-1& -1& -1& -1\\ -2& -2& -2& -2\end{array}\right]$ .

Add the two matrices:

  $\left[\begin{array}{cccc}-1& 3& 3& -1\\ 3& 3& -1& -1\end{array}\right]+\left[\begin{array}{cccc}-1& -1& -1& -1\\ -2& -2& -2& -2\end{array}\right]=\left[\begin{array}{cccc}\begin{array}{l}{A}^{\prime }\\ -2\end{array}& \begin{array}{l}{B}^{\prime }\\ 2\end{array}& \begin{array}{l}{C}^{\prime }\\ 2\end{array}& \begin{array}{l}{D}^{\prime }\\ -2\end{array}\\ 1& 1& -3& -3\end{array}\right]$ .

Graph the points represent by the matrix.

Hence, $\left[\begin{array}{cccc}\begin{array}{l}{A}^{\prime }\\ -2\end{array}& \begin{array}{l}{B}^{\prime }\\ 2\end{array}& \begin{array}{l}{C}^{\prime }\\ 2\end{array}& \begin{array}{l}{D}^{\prime }\\ -2\end{array}\\ 1& 1& -3& -3\end{array}\right]$ .