If $A = \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}$,then the determinant of the matrix $(A^{2016} - 2A^{2015} - A^{2014})$ is

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:

Let $B=\begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix}$ and $A$ be a $2 \times 2$ matrix such that $AB^{-1}=A^{-1}$. If $BCB^{-1}=A$ and $C^4+\alpha C^2+\beta I=O$,then $2\beta-\alpha$ is equal to:

If $AA^T = I$ and $C$ is a skew-symmetric matrix,then $((A^T CA)^{50})^T$ is equal to

Let $A = \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}$,where $a, c \in \mathbb{R}$. If $A^3 = A$ and the positive value of $a$ belongs to the interval $(n-1, n]$,where $n \in \mathbb{N}$,then $n$ is equal to $..........$.

For a $3 \times 3$ matrix $A$,if $A(\operatorname{adj} A) = \begin{bmatrix} -10 & 0 & 0 \\ 0 & -10 & 2 \\ 0 & 0 & -10 \end{bmatrix}$,then the value of the determinant of $A$ is: