Question

Find the image of the point $(2,-4)$ under the transformations $(x,y)\to (x+3y,y-x)$ is $(a,b)$, find $b-a?$

Answer 1

Let $(x_1,y_1)$ be the image of the point $(x,y)$ under the given transformations. Then
$\{\begin{array}{c}{x}_1=x+3y=1.x+3.y\\ y_1=y-x=(-1).x+1.y\end{array}$
$\Rightarrow \left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]\left[\begin{array}{c}2\\ -4\end{array}\right]=\left[\begin{array}{cc}2& -12\\ -2& -4\end{array}\right]=\left[\begin{array}{c}-10\\ -6\end{array}\right]$
$\therefore$ The image is $(-10-6)$