Computing the Inverse of a two dimensional map? In order to find the inverse of the function $y = x^3$ where $y = f(x) = x^3$ we need $x = f^{-1}(y)$, which we compute it as $x = y^{\frac{1}{3}}$ so the inverse function. But how do I calculate the inverse map of the following map? $x \mapsto Ax +By + C$ and $y \mapsto Dx$ ?, where $A,B,C,D$ are real numbers. I was trying to visualize this in terms of matrices, $\begin{bmatrix}x \\\ y\end{bmatrix} \mapsto \begin{bmatrix} Ax + By + C \\\ Dx\end{bmatrix}$, may that open up some new insights? How can we guarantee the existence of the inverse for this two dimensional map?

$x \mapsto Ax +By + C$ and $y \mapsto Dx$ can be rewritten as :

$$\left(\begin{matrix} x' \\\ y' \\\ 1 \end{matrix} \right)=M\left(\begin{matrix} x \\\y\\\1\end{matrix} \right)$$

Where

$$M=\left(\begin{matrix} A & B & C \\\ D & 0 & 0 \\\ 0 & 0 &1\end{matrix} \right).$$

If $M$ is invertible, then $\left(\begin{matrix} x \\\y\\\1\end{matrix} \right)=M^{-1}\left(\begin{matrix} x' \\\ y' \\\ 1 \end{matrix} \right)$.

So existence of an inverse is guaranteed by invertibility of the subjacent matrix when you have a linear system.

Which sums up here to $D\
e 0 \
e B$.

* * *

Note that we added a row for the affine part. By doing so, we where able to express your function as a linear system. We could have written also :

$$\left(\begin{matrix} x' \\\ y' \\\ 0 \end{matrix} \right)=M'\left(\begin{matrix} x \\\y\\\1\end{matrix} \right)$$

But the invertibility would have been more difficult to express in term of matrices.