If $A = \begin{bmatrix} 1 & -3 & 2 \\ -2 & 1 & 3 \\ 3 & 2 & -1 \end{bmatrix}$,then $A^2 \operatorname{Adj} A = $ (in $I$)

If $\operatorname{adj}\begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{bmatrix}$, then $[a \quad b]$ is equal to

If $A=\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{bmatrix}$ and $A^{-1}=\frac{1}{2}\begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & 2y \\ 5 & -3 & 1 \end{bmatrix}$, then the point $(x, y)$ lies on the curve represented by the equation:

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 $(AB)^{-1}$.

If $A = \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix}$,then $\text{adj}(3A^2 + 12A)$ is equal to

If $A = \begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix}$,then $A + \operatorname{adj}(A)$ is: