If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, $C = \begin{bmatrix} 7 & 3 \\ 0 & 6 \end{bmatrix}$ and $AB = C$, then the inverse of matrix $B$ is

If $A$ is an invertible matrix of order $n$, then the determinant of $\operatorname{adj} A$ is equal to :

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

Let $X=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$,$Y=\alpha I+\beta X+\gamma X^{2}$ and $Z=\alpha^{2} I-\alpha \beta X+\left(\beta^{2}-\alpha \gamma\right) X^{2}$,where $\alpha, \beta, \gamma \in \mathbb{R}$. If $Y^{-1}=\begin{bmatrix} \frac{1}{5} & \frac{-2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{-2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$,then $(\alpha-\beta+\gamma)^{2}$ is equal to

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 $A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix}$,then $(AB)^{-1} =$