If a matrix $A$ is such that $3A^3 + 2A^2 + 5A + I = 0$,then its inverse is

If $A = \frac{1}{5! 6! 7!} \begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}$,then $|\operatorname{adj}(\operatorname{adj}(2A))|$ is equal to:

If $A$ is a non-zero square matrix of order $n$ with $\det(I+A) \neq 0$ and $A^3=O$,where $I$ and $O$ are the identity and null matrices of order $n \times n$ respectively,then $(I+A)^{-1}$ is equal to

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

Assertion $(A)$: If $B$ is a $3 \times 3$ matrix and $|B|=6$,then $|\operatorname{Adj}(B)|=36$.
Reason $(R)$: If $B$ is a square matrix of order $n$,then $|\operatorname{Adj}(B)|=|B|^{n}$.

If $S = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$ and $A = \frac{1}{2} \begin{bmatrix} b+c & c-a & b-a \\ c-b & c+a & a-b \\ b-c & a-c & a+b \end{bmatrix}$,then $SAS^{-1} =$