If $A$ and $B$ are square matrices of the same order and $AB = 3I$, then $A^{-1}$ is equal to

If $A$ is a square matrix of order $3$ and $|A| = 8$,then $|adj(A)| = $

If $A$ and $B$ are square matrices of the same order such that $(A+B)(A-B)=A^{2}-B^{2}$,then $(A B A^{-1})^{2}$ is equal to

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

Let $A=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]$,$x \in R^{+}$ and $A^4=\left[a_{ij}\right]_2$. If $a_{11}=109$,then $\left(A^4\right)^{-1}=$

Matrix $A$ is a non-singular matrix and $(A-3I)(A-5I)=0$. Then,$\frac{15}{8} A^{-1} =$ . . . . . .