I am studying pure maths as a hobby. I am trying to understand inverse transformations. The text book I am using gives an example of how to do it:

Find the inverse transformation $T^{-1}$ of $T:\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto \begin{pmatrix}x\\x+y\\z\end{pmatrix}$

Let $\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\end{pmatrix}=\begin{pmatrix}x\\x+y\\z\end{pmatrix}$

Then: $x=p_{1}, z=p_{3}, y=p_{2}-x=p_{2}-p_{1}$

So $T^{-1}: \begin{pmatrix}p_{1}\\p_{2}\\p_{3}\end{pmatrix}\mapsto \begin{pmatrix}p_{1}\\

p_{2}-p_{1}\\

p_{3}\end{pmatrix}$

or $T^{-1}: \begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto \begin{pmatrix}x\\

y-x\\

z\end{pmatrix}$

An end of section question asks: Find $T^{-1}$ in the case where $T: \begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto \begin{pmatrix}x+y\\

x+y+z\\

2y-z\end{pmatrix}$

But I cannot successfully apply this technique. I have said:

$\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\end{pmatrix}=\begin{pmatrix}x+y\\x+y+z\\2y-z\end{pmatrix}$

$x+y=p_{1}, x+y+z=p_{2}, 2y-z=p_{3}$

Presumably I am meant to find a RHS consisting entirely of $p_{1},p_{2}$ and $p_{3}$ but I cannot find a solution to this.

## Best Answer

You have the system$$\left\{\begin{array}{l}x+y=p_1\\x+y+z=p_2\\2y-z=p_3.\end{array}\right.$$If you replace the second equation by the second equation minus the first one, it becomes$$\left\{\begin{array}{l}x+y=p_1\\z=p_2-p_1\\2y-z=p_3.\end{array}\right.$$So, now you know that $z=p_2-p_1$. It follows now from the third equation that $y=\frac12(-p_1+p_2+p_3)$. And now you can use the first equation to get $x$.