Linear Alg. Short proof on determinant Hi can I get a quick check on my proof to see if it is correct.
proof

 A: We are assumed to know that the inverse of a matrix $A$ exists iff $\det A\ne 0$ (assumed knowledge which I infer from StillLearning's proposed solution). We look at the simultaneous equations $ax+by=e$, $cx+dy=f$, and want to know when they have a solution $x, y$ for every $e$ and $f$.  We show that this happens iff the matrix 
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$
is invertible.  Based on our assumption, that proves that the equations are solvable iff the determinant is nonzero.
Based on the definition of matrix multiplication, we know that the simultaneous equations given are equivalent to $$\begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}e\\f\end{bmatrix}$$
Call the matrix $A$.  Multiply on both sides by the inverse of the matrix. Applying $A^{-1}A=I$, $I$ being the identity matrix, we get that $[x\,y]=A^{-1}[e\, f]$, so it follows that the existence of the inverse implies that the equations are solvable.  
To get the implication in the other direction (equations solvable implies inverse exists), let $x_1, y_1$ be the solution of the simultaneous equations for $e=1, f=0$, and let $x_2, y_2$ be the solution for $e=0, f=1$.  We have
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x_1&x_2\\y_1&y_2\end{bmatrix} = \begin{bmatrix}ax_1+by_1&ax_2+by_2\\cx_1+dy_1&cx_2+dy_2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$
The latter matrix is the identity, so the inverse exists and equals
$$\begin{bmatrix}x_1&x_2\\y_1&y_2\end{bmatrix}.$$
