If $A$ is a $3 \times 3$ matrix and $|A|=2$,then $|\operatorname{Adj}(\operatorname{Adj} A)| \operatorname{Adj}(\operatorname{Adj} A) = $ (in $A$)

Find the adjoint of the matrix $A = \left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$.

Let $A = (a_{ij})_{1 \leq i, j \leq 3}$ be a $3 \times 3$ invertible matrix where each $a_{ij}$ is a real number. Denote the inverse of the matrix $A$ by $A^{-1}$. If $\sum_{j=1}^3 a_{ij} = 1$ for $1 \leq i \leq 3$,then:

Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $\det(A)=-4$ and $A+I=\begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$,where $I$ is the identity matrix of order $3 \times 3$. If $\det((a+1) \operatorname{adj}((a-1) A)) = 2^m 3^n$,where $m, n \in \{0, 1, 2, \ldots, 20\}$,then $m+n$ is equal to:

If $A=\begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$,then $(\operatorname{Adj} A)^{-1}=$

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