If $\frac{x^2+7}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$, then the determinant of the matrix $\begin{bmatrix} A & B \\ C & \frac{2}{5} \end{bmatrix}$ is

Let $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$. Let $\alpha, \beta \in \mathbb{R}$ be such that $\alpha A^{2} + \beta A = 2I$. Then $\alpha + \beta$ is equal to -

Let $A = \begin{bmatrix} 2 & -5 \\ 3 & 1 \end{bmatrix}$. What is $f(A)$ if $f(x) = x^3 - 2x^2 - 5$?

Let $A$ be a $3 \times 3$ matrix such that $X^{T}AX = O$ for all nonzero $3 \times 1$ matrices $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$. If $A \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}1 \\ 4 \\ -5\end{array}\right]$,$A \left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]=\left[\begin{array}{l}0 \\ 4 \\ -8\end{array}\right]$,and $\operatorname{det}(\operatorname{adj}(2(A+I)))=2^\alpha 3^\beta 5^\gamma$,where $\alpha, \beta, \gamma \in \mathbb{N}$,then $\alpha^2+\beta^2+\gamma^2$ is

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = \alpha A + I$,where $\alpha \in R - \{-1, 1\}$. If $\det(A^2 - A) = 4$,then the sum of all possible values of $\alpha$ is equal to

Let $B=\begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$ and $C=\begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}$. If a matrix $A$ is such that $BAC=I$,then $A^{-1}=$