Answer
$AB^{-1}=-\frac{1}{320}\begin{bmatrix}
-17 & -8\\
-6 &-16
\end{bmatrix} $
Work Step by Step
We found in exercise 5 that $AB=\begin{bmatrix}
16 & 8\\
6 & -17
\end{bmatrix}$
Find an inverse matrix by obtaining augmented matrix:
$\begin{bmatrix}
16 & 8 | 1 & 0\\
6 & -17 | 0 & 1
\end{bmatrix} \approx^1 \begin{bmatrix}
1 & \frac{1}{2} | \frac{1}{16} & 0\\
6 & -17 | 0 & 1
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & \frac{1}{2} | \frac{1}{16} & 0\\
0 & -20 | -\frac{3}{8} & 1
\end{bmatrix} \approx^3 \begin{bmatrix}
1 & \frac{1}{2} | \frac{1}{16} & 0\\
0 & 1 | \frac{3}{160} & -\frac{1}{20}
\end{bmatrix} \approx^4 \begin{bmatrix}
1 & 0| \frac{17}{320} & \frac{1}{40}\\
0 & 1 | \frac{3}{160} & -\frac{1}{20}
\end{bmatrix}$
$1.M_1(\frac{1}{16})$
$2.A_{12}(-6)$
$3.M_2(-\frac{1}{20})$
$4.A_{21}(-\frac{1}{2})$
Hence here, $AB^{-1}=\begin{bmatrix}
\frac{17}{320} & \frac{1}{40}\\
\frac{3}{160} & -\frac{1}{20}
\end{bmatrix} =-\frac{1}{320}\begin{bmatrix}
-17 & -8\\
-6 &-16
\end{bmatrix} $
