If $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$ and $A + A^{-1} = I$,then $\alpha =$

If $A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix}$,then $(AB)^{-1} =$

If $(BA)^{-1} = C$ where $B = \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$ and $C = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}$,then $A^{-1}$ is given by

If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \operatorname{adj} A = A^T$,then $5a + b$ is equal to

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 $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(AB)^{-1} = B^{-1} A^{-1}$.