Let $\mathrm{A}=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha>0$, such that $\operatorname{det}(\mathrm{A})=0$ and $\alpha+\beta=1$. If I denotes $2 \times 2$ identity matrix, then the matrix $(I+A)^8$ is :

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: