If $A^2-A+I=0$, then the inverse of the matrix $A$ is

Let $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$ and $A^{-1} = xA + yI_2$,(where $I_2$ is the unit matrix of order $2$),then

If $A = \begin{bmatrix} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{bmatrix}$ and $AB = I$,then $B$ is equal to

Find the inverse of the matrix (if it exists): $\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$

Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=12^4$. Then $|A^{-1} \operatorname{adj} A|$ is equal to

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$,such that $A^{2} - 4A + 3I = 0$,then $A^{-1} =$