## Precalculus (6th Edition) Blitzer

Let the given $2\times2$ matrix be $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ We can find its multiplicative inverse by switching the elements $a,d$, adding negative signs to $b,c$, and then dividing the new matrix by a factor of $(ad-bc)$. That is, $A^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$