Chapter 2.4, Problem 21E

To determine

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

Answer to Problem 21E

  $H^{\prime }\left(-2,-1\right),I^{\prime }\left(1,-3\right),J^{\prime }\left(5,-1\right)\text{ and }{K}^{\prime }\left(4,2\right)$ .

Explanation of Solution

Given:

A trapezoid with vertex:

  $H\left(-1,-2\right),I\left(-3,1\right),J\left(-1,5\right)\text{ and }K\left(2,4\right)$ .

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:

A trapezoid with vertex:

  $H\left(-1,-2\right),I\left(-3,1\right),J\left(-1,5\right)\text{ and }K\left(2,4\right)$ .

First write the vertex matrix as:

  $\left[\begin{array}{cccc}\begin{array}{l}H\\ -1\end{array}& \begin{array}{l}I\\ -3\end{array}& \begin{array}{l}J\\ -1\end{array}& \begin{array}{l}K\\ 2\end{array}\\ -2& 1& 5& 4\end{array}\right]$ .

Reflected over the $y=x$ and then $R_{y=x}$ is:

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

Multiply by the $y$ -axis reflection matrix.

  $$

  $\begin{array}{l}\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cccc}-1& -3& -1& 2\\ -2& 1& 5& 4\end{array}\right]\\ \Rightarrow \left[\begin{array}{cccc}\begin{array}{l}{H}^{\prime }\\ -2\end{array}& \begin{array}{l}{I}^{\prime }\\ 1\end{array}& \begin{array}{l}{J}^{\prime }\\ 5\end{array}& \begin{array}{l}{K}^{\prime }\\ 4\end{array}\\ -1& -3& -1& 2\end{array}\right].\end{array}$

Graph of the two trapezoids is:

Hence, the vertices of the image are $H^{\prime }\left(-2,-1\right),I^{\prime }\left(1,-3\right),J^{\prime }\left(5,-1\right)\text{ and }{K}^{\prime }\left(4,2\right)$ .