If $A$ and $B$ are square matrices of the same order such that $(A+B)(A-B)=A^{2}-B^{2}$,then $(A B A^{-1})^{2}$ is equal to

If $A = \begin{bmatrix} 1 & \cot \frac{\theta}{2} \\ -\cot \frac{\theta}{2} & 1 \end{bmatrix}$,then $A^{-1} =$

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 $\sum_{1 \leq i, j \leq 3} a_{ij} =$

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:

Find the inverse of the matrix (if it exists): $\left[\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$

Let $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that $[adj A]^{-1} = adj(A^{-1})$.