If $A = \begin{bmatrix} a & 1 & 2 \\ 1 & 2 & b \\ c & 1 & 3 \end{bmatrix}$ and $\operatorname{Adj} A = \begin{bmatrix} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{bmatrix}$,then $a^2 + b^2 + c^2 = $

If $A=\begin{bmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix}$ and $A^{-1}=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, then $\sum_{1 \leq i, j \leq 3} a_{ij} =$

If $A=\begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$,then $(\operatorname{Adj} A)^{-1}=$

Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=81$. If $S =\{ n \in \mathbb{Z} :(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^2}{2}}=|A|^{(3n^2-5n-4)}\}$,then $\sum_{n \in S}|A^{(n^2+n)}|$ is equal to

Let $A = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}$,where $i = \sqrt{-1}$. If $M = A^{T}BA$,then the inverse of the matrix $AM^{2023}A^{T}$ is $.........$

Which of the following is true for the matrix product $AB$?