# How to do this question 9 regarding matrices and transformations?

Feb 12, 2018

$\textcolor{b l u e}{y = - \frac{1}{3} x + 3}$

#### Explanation:

All points on the line $y = 2 x + 3$ will be transformed by the matrix $\left(\begin{matrix}0 & 3 \\ - 2 & 0\end{matrix}\right)$

Any point on the line will have coordinates of the form. $\left(k , 2 k + 3\right)$

Thus:

$\left(\begin{matrix}x ' \\ y '\end{matrix}\right) = \left(\begin{matrix}0 & 3 \\ - 2 & 0\end{matrix}\right) \left(\begin{matrix}k \\ 2 k + 3\end{matrix}\right)$

$\textcolor{w h i t e}{888888} = \left(\begin{matrix}0 + 6 k + 9 \\ - 2 k + 0\end{matrix}\right) = \left(\begin{matrix}6 k + 9 \\ - 2 k\end{matrix}\right)$

i.e. $x ' = 6 k + 9$ and $y ' = - 2 k$

Eliminating $k$

$k = \frac{x - 9}{6}$

$y = - 2 \left(\frac{x - 9}{6}\right)$

$\textcolor{b l u e}{y = - \frac{1}{3} x + 3}$

$y = - \frac{1}{3} x + 3$ is the image of $y = 2 x + 3$ under the transformation:

$\left(\begin{matrix}0 & 3 \\ - 2 & 0\end{matrix}\right)$

Method 2

Since we have:

$\boldsymbol{A X} = \boldsymbol{{X}^{'}}$

Where $\boldsymbol{A} = \left(\begin{matrix}0 & 3 \\ - 2 & 0\end{matrix}\right)$

$\boldsymbol{X} = \left(\begin{matrix}x \\ y\end{matrix}\right)$ and $\boldsymbol{{X}^{'}} = \left(\begin{matrix}x ' \\ y '\end{matrix}\right)$

Generating a pair of coordinates using the line $y = 2 x + 3$

$x = 3$ and $x = 4$

gives:

$y = 9$ and $y = 11$ respectively.

Putting these under the transformation:

$\left(\begin{matrix}0 & 3 \\ - 2 & 0\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}x ' \\ y '\end{matrix}\right)$

$\therefore$

$\left(\begin{matrix}0 & 3 \\ - 2 & 0\end{matrix}\right) \left(\begin{matrix}3 \\ 9\end{matrix}\right) = \left(\begin{matrix}27 \\ - 6\end{matrix}\right)$

$\left(\begin{matrix}0 & 3 \\ - 2 & 0\end{matrix}\right) \left(\begin{matrix}4 \\ 11\end{matrix}\right) = \left(\begin{matrix}33 \\ - 8\end{matrix}\right)$

$\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{- 8 - \left(- 6\right)}{33 - 27} = \frac{- 2}{6} = - \frac{1}{3}$

Using point slope form of a line:

$\left({y}_{2} - {y}_{1}\right) = m \left({x}_{2} - {x}_{1}\right)$ $\textcolor{w h i t e}{88} m = \text{gradient}$

$y - \left(- 8\right) = - \frac{1}{3} \left(x - 33\right)$

$y + 8 = - \frac{1}{3} x + 11$

$y = - \frac{1}{3} x + 3 \textcolor{w h i t e}{888}$ as expected