$$\text {If}, \mathrm{A}=\left[\begin{array}{cc} 2 & 3\\ -1 & 0 \end{array}\right];\quad \mathrm{B}=\left[\begin{array}{cc} -1 & 2\\ 0 & 1 \end{array}\right] \text {show that }(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}$$

Solution:

$\begin{aligned} \mathrm{AB} &=\left[\begin{array}{cc} 2 & 3 \\\\ -1 & 0 \end{array}\right]\left[\begin{array}{cc} -1 & 2 \\\\ 0 & 1 \end{array}\right] \\\\ \mathrm{AB} &=\left[\begin{array}{cc} -2 & 7 \\\\ 1 & -2 \end{array}\right] \\\\ |\mathrm{AB}| &=\left[\begin{array}{cc} -2 & 7 \\\\ 1 & -2 \end{array}\right]=-3 \neq 0 \end{aligned} \\ $

$(A B)^{-1}$ …