if find find $(A B)^{-1}$

We know that $(A B)^{-1}=B^{-1} A^{-1}$.

$B=\left[\begin{array}{rrc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$

$\therefore|B|=1 \times 3-2 \times(-1)-2(2)=3+2-4=5-4=1$

Now, $A_{11}=3, A_{12}=1, A_{13}=2$

$A_{21}=2, A_{22}=1, A_{23}=2$

$A_{31}=6, A_{3,}=2, A_{33}=5$

$\therefore \operatorname{adj} B=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$

Now,

$B^{-1}=\frac{1}{|B|} \cdot a d j B$

$\therefore B^{-1}=\left[\begin{array}{ccc}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$

$\therefore(A B)^{-1}=B^{-1} A^{-1}$

$=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]\left[\begin{array}{crr}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$

$=\left[\begin{array}{lll}9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10\end{array}\right]$

$=\left[\begin{array}{ccc}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{array}\right]$