If $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right],$ find $(AB)^{-1}$.

(D) We know that $(AB)^{-1} = B^{-1}A^{-1}$.
First,we calculate $B^{-1}$. We know that $B^{-1} = \frac{1}{|B|} \text{adj}(B)$,which exists if $|B| \neq 0$.
Calculating $|B|$:
$|B| = \left|\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right| = 1(3-0) - 2(-1-0) - 2(2-0) = 3 + 2 - 4 = 1$.
Since $|B| = 1 \neq 0$,$B^{-1}$ exists.
Calculating $\text{adj}(B)$:
The matrix of cofactors is obtained by calculating the cofactor $A_{ij}$ for each element.
$A_{11} = 3, A_{12} = 1, A_{13} = 2$
$A_{21} = 2, A_{22} = 1, A_{23} = 2$
$A_{31} = 6, A_{32} = 2, A_{33} = 5$
Thus,$\text{adj}(B) = \left[\begin{array}{ccc}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$.
Since $|B| = 1$,$B^{-1} = \text{adj}(B) = \left[\begin{array}{ccc}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$.
Now,$(AB)^{-1} = B^{-1}A^{-1} = \left[\begin{array}{ccc}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right] \left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$.
Performing matrix multiplication:
$(AB)^{-1} = \left[\begin{array}{ccc}9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10\end{array}\right] = \left[\begin{array}{ccc}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{array}\right]$.