The element in the third row and first column of the inverse of the matrix $\left[\begin{array}{ccc}1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]$ is

$\begin{aligned} & A(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^\beta\end{array}\right] \\ & \Rightarrow[A(\alpha, \beta)]^{-1}=\end{aligned}$

If $A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$,then $A(\text{adj } A) = $

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

If $A = \begin{bmatrix} k & 5 & 2 \\ 2 & -k & 5 \\ 5 & 2 & -k \end{bmatrix}$ and $\det A = 190$, then $\operatorname{Adj} A = $

If $A = \begin{bmatrix} a+ib & c+id \\ -c+id & a-ib \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} a+ib & -c-id \\ -c+id & a-ib \end{bmatrix}$,find $(a^2+b^2+c^2+d^2)$.