Show that (under appropriate assumptions) a general linear mapping $F(x,y) = (ax+by,cx+dy)$ is invertible over all of $\Bbb R^2$ (i.e. there is a single inverse for all of $\Bbb R^2$). What conditions must you assume on $a,b,c,d$?
The derivative matrix is:
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix},$$ and the determinant is $ad-bc$.
My answer is that it is invertible as long as $ad \neq bc$ (so that the determinant is not zero).
Is this a complete answer?
It is best to set it out in all of its detail. Let $\mathrm{f} : \mathbb{R}^2 \to \mathbb{R}^2$ be linear, with $(x,y)$ as coordinates in the domain and $(u,v)$ as coordinates in the range. We have $u=ax+by$ and $v=cx+dy$: $$\begin{eqnarray*} ax+by&=& u \\ cx+dy &=&v \end{eqnarray*}$$ This can be re-written as a single matrix equation: $$\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{c} u \\ v \end{array}\right]$$ This system of linear equations has a unique solution if, and only if, the $2\times 2$ matrix on the left-hand-side is non-singular, i.e. if, and only if, $ad-bc \neq 0$. If $ad-bc \neq 0$ then $$\left[\begin{array}{c} x \\ y \end{array}\right] = \frac{1}{ad-bc}\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]\left[\begin{array}{c} u \\ v \end{array}\right]$$