If $A = \begin{bmatrix} 1 & 2 & i \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix}$,then $[\operatorname{adj}(\operatorname{adj} A)]^{-1} = $

Let $A = (a_{ij})_{1 \leq i, j \leq 3}$ be a $3 \times 3$ invertible matrix where each $a_{ij}$ is a real number. Denote the inverse of the matrix $A$ by $A^{-1}$. If $\sum_{j=1}^3 a_{ij} = 1$ for $1 \leq i \leq 3$,then:

Let the matrix $A = \begin{bmatrix} 10^{30} + 5 & 10^{20} + 4 & 10^{20} + 6 \\ 10^4 + 2 & 10^8 + 7 & 10^{10} + 2n \\ 10^4 + 8 & 10^6 + 4 & 10^{15} + 9 \end{bmatrix}$,where $n \in N$. Then:

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$,then $A^{-1}$ is given by

If $AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$ and $11B^{-1} = \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix}$,then $A = $ . . . . . . .

If $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right],$ find $(AB)^{-1}$.