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

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} \\ $

$(A B)^{-1}$ exists,

$ \begin{aligned} \operatorname{Adj}(\mathrm{AB}) &=\left(\begin{array}{rr} -2 & -7 \\\\ -1 & -2 \end{array}\right) \\\\ (\mathrm{AB})^{-1} &=\frac{1}{-3}\left(\begin{array}{rr} -2 & -7 \\\\ -1 & -2 \end{array}\right) \\\\ (\mathrm{AB})^{-1} &=\frac{-1}{3}\left(\begin{array}{rr} -2 & -7 \\\\ -1 & -2 \end{array}\right)........(1) \end{aligned} $

$ B=\left(\begin{array}{cc} -1 & 2 \\\\ 0 & 1 \end{array}\right) \quad|B|=\left|\begin{array}{cc} -1 & 2 \\\\ 0 & 1 \end{array}\right|=-1 \neq 0 $

$ \operatorname{Adj} \mathrm{B}=\left(\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right) \\ $

$ \begin{aligned} \mathrm{B}^{-1} &=\frac{1}{-1}\left(\begin{array}{cc} -1 & 2 \\\ 0 & 1 \end{array}\right) \\\\ \mathrm{A} &=\left[\begin{array}{cc} 2 & 3 \\\\ -1 & 0 \end{array}\right] \quad|\mathrm{A}=| \begin{array}{cc} 2 & 3 \\\\ -1 & 0 \end{array} \mid=3 \neq 0 \end{aligned} \\ $

$ \operatorname{Adj} \mathrm{B}=\left(\begin{array}{cc}-1 & 2 \\\\ 0 & 1\end{array}\right) \\ $

$ \mathrm{B}^{-1}=\frac{1}{-1}\left(\begin{array}{cc}-1 & 2 \\\\ 0 & 1\end{array}\right) \\ $

$ \mathrm{A}=\left[\begin{array}{cc}2 & 3 \\\\ -1 & 0\end{array}\right] \quad|\mathrm{A} \models| \begin{array}{cc}2 & 3 \\\\ -1 & 0\end{array} \mid=3 \neq 0 \\ $

$ \operatorname{Adj} \mathrm{A}=\left[\begin{array}{cc}0 & -3 \\\\ 1 & 2\end{array}\right] \\ $

$ \mathrm{A}^{-1}=\frac{1}{3}\left[\begin{array}{cc}0 & -3 \\\\ 1 & 2\end{array}\right] \\ $

$ \mathrm{B}^{-1} \mathrm{~A}^{-1}=-\frac{1}{1}\left[\begin{array}{cc}1 & -2 \\\\ 0 & -1\end{array}\right] \times \frac{1}{3}\left[\begin{array}{cc}0 & -3 \\\\ 1 & 2\end{array}\right] \\ $

$ \mathrm{B}^{-1} \mathrm{~A}^{-1}=-\frac{1}{3}\left[\begin{array}{cc}1 & -2 \\\\ 0 & -1\end{array}\right]\left[\begin{array}{cc}0 & -3 \\\\ 1 & 2\end{array}\right] \\ $

$ \mathrm{B}^{-1} \mathrm{~A}^{-1}=-\frac{1}{3}\left[\begin{array}{ll}-2 & -7 \\\\ -1 & -2\end{array}\right]........(2) \\ $

From (i) and (ii), $ (\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1} \\ $