If $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$,then the inverse of the matrix $A^3$ is

Assertion $(A)$: If $B$ is a $3 \times 3$ matrix and $|B|=6$,then $|\operatorname{Adj}(B)|=36$.
Reason $(R)$: If $B$ is a square matrix of order $n$,then $|\operatorname{Adj}(B)|=|B|^{n}$.

$\begin{aligned} & A(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^\beta\end{array}\right] \\ & \Rightarrow[A(\alpha, \beta)]^{-1}=\end{aligned}$

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 $(AB)^{-1}$.

Let $A$ be a $n \times n$ matrix such that $|A|=2$. If the determinant of the matrix $\operatorname{Adj}(2 \cdot \operatorname{Adj}(2A^{-1}))$ is $2^{84}$,then $n$ is equal to:

${\left[ {\begin{array}{*{20}{c}}{ - 6}&5\\{ - 7}&6\end{array}} \right]^{ - 1}}$ =