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

The inverse of matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is

Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = \begin{bmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{bmatrix}$,$|A| = -1$. Let $B$ be the inverse of the matrix $\operatorname{adj}(A \operatorname{adj}(A^2))$. Then $|\lambda B + I|$ is equal to . . . . . . .

If $A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 \\ 4 & -1 \end{bmatrix}$,then $(A+B)^{-1}$ is

Find $adj$ $A$ for $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.

If $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$,then