For $\alpha, \beta \in R$ and a natural number $n$,let $A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$. Then $2A_{10} - A_8$ is equal to:

The number of integers $x$ satisfying $-3 x^4 + \operatorname{det}\begin{bmatrix} 1 & x & x^2 \\ 1 & x^2 & x^4 \\ 1 & x^3 & x^6 \end{bmatrix} = 0$ is equal to

Let $S = \left\{ \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix} : a, b \in \{1, 2, 3, \ldots, 100\} \right\}$ and let $T_n = \{A \in S : A^{n(n+1)} = I\}$. Then the number of elements in $\bigcap_{n=1}^{100} T_n$ is

If $A$ is a square matrix such that $A^{2} = A$,then $(I + A)^{3} - 7A$ is equal to

Let three matrices $A = \begin{bmatrix} 2 & 1 \\ 4 & 1 \end{bmatrix}$,$B = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix}$,and $C = \begin{bmatrix} 3 & -4 \\ -2 & 3 \end{bmatrix}$. Then find the value of $Tr(A) + Tr\left( \frac{ABC}{2} \right) + Tr\left( \frac{A(BC)^2}{4} \right) + Tr\left( \frac{A(BC)^3}{8} \right) + \dots + \infty$.

Let $a = \lim_{x \to 1} \left( \frac{x}{\ln x} - \frac{1}{x \ln x} \right)$,$b = \lim_{x \to 0} \frac{x^3 - 16x}{4x + x^2}$,$c = \lim_{x \to 0} \frac{\ln(1 + \sin x)}{x}$,and $d = \lim_{x \to -1} \frac{(x + 1)^3}{3(\sin(x + 1) - (x + 1))}$. Then the matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is: