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?
 A: $x \mapsto Ax +By + C$ 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 $D\ne 0 \ne B$.

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.
A: So you have the mapping $f\colon \mathbb{R}^2\to  \mathbb{R}^2$, $f(x,y)=(Ax+By+C,Dx)$. Claim: $f$ has an inverse iff $A,B\not=0$. This follows from the fact that $f$ is invertible iff the linear system of equations $Ax+By=\xi-C, Dx= \eta$ is solvable for all $(\xi,\eta)\in  \mathbb{R}^2$.
Missprint: The condition is $B,D\not=0$
