If $A=\begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}$ and $A^{2}-5A-6I=0$,then $A^{-1}=$

The inverse of a symmetric matrix is

If $A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$,then $A \cdot \operatorname{adj}(A)$ is equal to

If $A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$,then $A^{-1} =$

If $A$ is a $2 \times 2$ order non-singular matrix,then the determinant of $A^{-1}$ is . . . . . . .

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 ...