If $A$ is a non-singular matrix and $A^2 - A + I = 0$, then $A^{-1} = \dots$

If $A$ and $B$ are non-singular matrices and $\operatorname{det}(AB)=(\operatorname{det} A)(\operatorname{det} B)$,then $((\operatorname{det} A)(\operatorname{det} B)) B^{-1} A^{-1} =$

If the inverse matrix of $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$ is $A^{-1} = \begin{bmatrix} a & 3/11 \\ 1/11 & b \end{bmatrix}$, then $a+b=$ . . . . . . .

If $A$ is a square matrix of order $4$ and $B = \text{Adj}(A)$,where $|B| = 27$,then the value of $|A^{-1} \text{Adj}(3AB)|$ is,(where $A^{-1}$ denotes the inverse of matrix $A$ and $\text{Adj}(A)$ denotes the adjoint of matrix $A$):

If $A=\left[\begin{array}{ccc}1 & 2 & 1 \\ -1 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 2 \\ -3 & 1 \\ 0 & 2\end{array}\right]$,then find $(AB)^{-1}$.

If $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right]$ and $\operatorname{adj} A=\left[\begin{array}{ccc}4 & -1 & 1 \\ 8 & -7 & a \\ 9 & -6 & b\end{array}\right]$,then find the values of $a$ and $b$.