Chapter 2.4, Problem 22E

To determine

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

Answer to Problem 22E

The vertices of the image are $L^{\prime }\left(1,1\right),M^{\prime }\left(2,2\right)\text{ and }{N}^{\prime }\left(1,3\right)$ .

Explanation of Solution

Given:

Triangle $LMN$ with vertices:

  $L\left(1,-1\right),M\left(2,-2\right)\text{ and }N\left(3,-1\right)$ .

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:

Triangle $LMN$ with vertices:

  $L\left(1,-1\right),M\left(2,-2\right)\text{ and }N\left(3,-1\right)$ .

First write the vertex matrix as:

  $\left[\begin{array}{ccc}\begin{array}{l}L\\ 1\end{array}& \begin{array}{l}M\\ 2\end{array}& \begin{array}{l}N\\ 3\end{array}\\ -1& -2& -1\end{array}\right]$ .

Counter clockwise ${90}^0$ rotation about the origin.

  $Ro{t}_{{90}^0}$ is:

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

Then multiply it by vertex matrix:

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

The graph of the two triangles is:

Hence, the vertices of the image are $L^{\prime }\left(1,1\right),M^{\prime }\left(2,2\right)\text{ and }{N}^{\prime }\left(1,3\right)$ .