Question

Find the image of the point $(2,-4)$ under the transformations $(x,y)$ $\to (x+3y,y-x)$. Let it be $(a,b)$. Then $b-a$ is

Answer 1

Let $(x_1,y_1)$ be the image of the point $(x,y)$. Then by the given transformation, we have

$x_1=1\times x+3\times y$

$y_1=(-1)x+1\times y$

$\Rightarrow \left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]\left[\begin{array}{c}2\\ -4\end{array}\right]=\left[\begin{array}{c}-10\\ -6\end{array}\right]$

Therefore, the image is $(-10,-6).$

So, $b-a=4$