## Precalculus (6th Edition) Blitzer

The inverse of a $2\times 2$ matrix is determined as below: $A=\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]$ $B=\left[ \begin{matrix} e & f \\ g & h \\ \end{matrix} \right]$ Now, we will consider that the multiplication equals, $\left[ AB \right]=\left[ \begin{matrix} p & q \\ r & s \\ \end{matrix} \right]$ Then, the inverse is ${{\left[ AB \right]}^{-1}}=\frac{1}{ps-qr}\left[ \begin{matrix} s & -r \\ -q & p \\ \end{matrix} \right]$ Thus, the matrix is invertible. If $ad-bc\ne 0$, then it is not.