If $A$ is a square matrix of order $3 \times 3$ and $\operatorname{det}(A) = 3$,then the value of $\operatorname{det}(3A^{-1})$ is:

Find $adj$ $A$ for $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.

If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}_{3 \times 3}$, and $A^{-1} = \begin{bmatrix} \gamma & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}_{3 \times 3}$, then $|\alpha \cdot \beta \cdot \gamma| = $ (where $| \cdot |$ denotes the absolute value)

If matrix $A$ and its inverse $A^{-1}$ are given by $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & y \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}$, then the polar coordinates of the point whose Cartesian coordinates are $(x, y)$ are ...

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 $\text{adj}(3A^2 + 12A) = \dots$