If $A = \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix}$ where $i = \sqrt{-1}$ and $B = A^{2029}$,then $B^{-1} =$

Let $A$ be a $3 \times 3$ matrix such that $A+A^{T}=O$. If $A\begin{bmatrix}1\\ -1\\ 0\end{bmatrix}=\begin{bmatrix}3\\ 3\\ 2\end{bmatrix}$, $A^{2}\begin{bmatrix}1\\ -1\\ 0\end{bmatrix}=\begin{bmatrix}-3\\ 19\\ -24\end{bmatrix}$ and $\det(\text{adj}(2\text{adj}(A+I))) = (2)^\alpha \cdot(3)^\beta \cdot(11)^\gamma$, then $\alpha+\beta+\gamma$ is equal to . . . . . . .

If $F(\alpha ) = \begin{bmatrix} \cos \alpha & - \sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $G(\beta ) = \begin{bmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ - \sin \beta & 0 & \cos \beta \end{bmatrix}$,then $[F(\alpha ) G(\beta )]^{-1} = $

Let $I$ be a unit matrix of order $6$. Let $A = (a_{ij})$ be a square matrix of order $6$ such that $a_{ij} = \begin{cases} 1, & \text{if } i+j=7 \\ 0, & \text{if } i+j \neq 7 \end{cases}$. Then $(A(\text{adj } A) A^{-1}) A^2 = $

If $A$ and $B$ are invertible matrices,then which of the following is not correct?

If $A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$, then $(A^2)^{-1} = $