Question
Let $\text{A}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}\text{and B}=\begin{bmatrix}6 & 7 \\8 & 9 \end{bmatrix}$. Find $(AB)^{-1}.$

Answer

$\text{A}=\begin{bmatrix}3 & 2\\7 & 5 \end{bmatrix}\therefore\ |\text{A}|=1\neq0\text{ and adj A}=\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$$\therefore\ \text{A}^{-1}\frac{\text{adj A}}{|\text{A}|}=\frac{1}{1}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$\text{B}=\begin{bmatrix}6 & 7 \\7 & 9 \end{bmatrix}\therefore\ |\text{B}|=-2\neq0\text{ and adj B}=\begin{bmatrix}9 & -7 \\-8 & 6 \end{bmatrix}$
$\therefore\ \text{B}^{-1}=\frac{\text{adj B}}{|\text{B}|}=\frac{1}{(-2)}=\begin{bmatrix}9 & -7 \\-8 & 6 \end{bmatrix}$
Now, $(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1}$
$(\text{AB})^{-1}=\frac{1}{(-2)}\begin{bmatrix}9 & -7 \\-8 & 6 \end{bmatrix}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$(\text{AB})^{-1}=-\frac{1}{2}\begin{bmatrix}94 & -39 \\-82 & 34 \end{bmatrix}$
$\text{(AB)}^{-1}=\begin{bmatrix}-47 & \frac{39}{2} \\41 & -17 \end{bmatrix}$

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