Inverse image of $P=\left \{ (x,y)\in\mathbb{R}^{2}\mid 0<x+y<1,0<2x-3y<4 \right \}$

Question

Considering $$P=\left \{ (x,y)\in\mathbb{R}^{2}\mid 0<x+y<1,0<2x-3y<4 \right \}$$ and the transformation $$\begin{pmatrix} x\\ y \end{pmatrix} \mapsto \begin{pmatrix} 1 & 1\\2 &-3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}$$ I'm looking for the inverse image of $P$. I suppose it has to be something like $$P^{-1}=\left \{ (x,y)\in\mathbb{R}^{2}\mid \alpha _{1}<x<\beta _{1},\alpha _{2}<y<\beta _{2} \right \}$$ but I don't know how to calculate the bounds $\alpha_{i}$ and $\beta_{i}$ for $i=1,2$. How I can find these?

Answer 1

When you apply the matrix transformation, what you are doing is transforming the coordinates $(x,y)$ into new coordinates $(u,v).$ In this case,

$$\begin{bmatrix} u \\ v\\ \end{bmatrix}=\begin{bmatrix} 1 && 1 \\ 2 && -3\\ \end{bmatrix}\begin{bmatrix} x \\ y\\ \end{bmatrix}$$

which means $u=x+y$ and $v=2x-3y.$ Since $0<x+y<1,$ this implies by the transformation $0<u<1$ and since $0<2x-3y<4$ this implies by the transformation that $0<v<4.$ So the inequalities are $0<u<1$ and $0<v<4.$