If a matrix $A$ satisfies the equation $A^3-6A^2+11A-6I=0$,then $A^{-1}$ can be expressed in terms of $A$ as:

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 $\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = I_2$,then $A =$

The inverse of the matrix $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$ is

If $A$ is a square matrix of order $3 \times 3$,then $|\operatorname{adj} A| =$ . . . . . . .

Matrix $A = \begin{bmatrix} 1 & 0 & -k \\ 2 & 1 & 3 \\ k & 0 & 1 \end{bmatrix}$ is invertible for