Considering $$P=\left \{ (x,y)\in\mathbb{R}^{2}\mid 0
Inverse image of $P=\left \{ (x,y)\in\mathbb{R}^{2}\mid 0
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functions
linear-transformations
inverse-function
1 Answers
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When you apply the matrix transformation, what you are doing is transforming the coordinates $(x,y)$ into new coordinates $(u,v).$ In this case,
$$\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
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0Well I calculated $\left ( 0,0 \right )$,$\left ( \frac{11}{25},\frac{4}{25} \right )$, $\left ( \frac{8}{25},\frac{12}{25} \right )$ and $\left ( \frac{19}{25},\frac{16}{25} \right )$ as the vertices of $P^{-1}$. But I still don't know the bounds. – 2017-02-04
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0I think those are the vertices of $P.$ Let me check. – 2017-02-04
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0I will update my answer because I think I found an easier way. – 2017-02-04
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0Ok I fixed my answer. Let me know if it is unclear still. – 2017-02-04