Chapter 2.4, Problem 5CFU

To determine

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

Answer to Problem 5CFU

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

Explanation of Solution

Given:

Triangle $JKL$ has vertices:

  $J\left(-2,5\right),K\left(1,3\right)\text{ and }L\left(0,-2\right)$ .

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:

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

Calculation:

Triangle $JKL$ has vertices:

  $J\left(-2,5\right),K\left(1,3\right)\text{ and }L\left(0,-2\right)$ .

First write the coordinate of the vertices as a matrix. Then do a scalar multiplication using the scalar factor.

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

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

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