Chapter 8.4, Problem 89AE

Consider the 2 by 2 square matrix $A\text{ = }\left[\begin{array}{c}a\\ c\end{array}\begin{array}{c}b\\ d\end{array}\right]$.

If $D\text{ = }ad-bc\ne \text{ 0}$, show that $A$ is nonsingular and that,

$A^{-1}\text{ = }\frac{1}{D}\left[\begin{array}{c}d\\ -c\end{array}\begin{array}{c}-b\\ a\end{array}\right]$