Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $\operatorname{det}(A)=-4$ and $A+I=\left[\begin{array}{lll}1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2\end{array}\right]$, where $I$ is the identity matrix of order $3 \times 3$. If $\operatorname{det}((a+1) \operatorname{adj}((a-1) A))$ is $2^{\mathrm{m}} 3^{\mathrm{n}}, \mathrm{m}$, $\mathrm{n} \in\{0,1,2, \ldots, 20\}$, then $\mathrm{m}+\mathrm{n}$ is equal to :

If the system of linear equations

$$ \begin{aligned} & 3 x+y+\beta z=3 \\ & 2 x+\alpha y-z=-3 \\ & x+2 y+z=4 \end{aligned} $$

has infinitely many solutions, then the value of $22 \beta-9 \alpha$ is :

Let $A = [a_{ij}]$ be a $2 \times 2$ matrix such that $a_{ij} \in \{0, 1\}$ for all $i$ and $j$. Let the random variable $X$ denote the possible values of the determinant of the matrix $A$. Then, the variance of $X$ is:

Let $ \alpha, \beta \ (\alpha \neq \beta) $ be the values of $ m $, for which the equations $ x+y+z=1 $, $ x+2y+4z=m $ and $ x+4y+10z=m^2 $ have infinitely many solutions. Then the value of $ \sum\limits_{n=1}^{10} (n^{\alpha}+n^{\beta}) $ is equal to :