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

If ${\Delta _1} = \left| {\begin{array}{*{20}{c}} x & {\sin \theta } & {\cos \theta } \\ {\sin \theta } & { - x} & 1 \\ {\cos \theta } & 1 & x \end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}} x & {\sin 2\theta } & {\cos 2\theta } \\ {\sin 2\theta } & { - x} & 1 \\ {\cos 2\theta } & 1 & x \end{array}} \right|$,$x \ne 0$; then for all $\theta \in \left( {0, \frac{\pi }{2}} \right)$:

Match the statements given in Column $I$ with the intervals/union of intervals given in Column $II$.

Column $I$	Column $II$
$(A)$ The set $\{\operatorname{Re}(\frac{2 i z}{1-z^2}): |z|=1, z \neq \pm 1\}$ is	$(p)$ $(-\infty,-1) \cup(1, \infty)$
$(B)$ The domain of $f(x)=\sin ^{-1}(\frac{8(3)^{x-2}}{1-3^{2(x-1)}})$ is	$(q)$ $(-\infty, 0) \cup(0, \infty)$
$(C)$ If $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$,then the set $\{f(\theta): 0 \leq \theta < \frac{\pi}{2}\}$ is	$(r)$ $[2, \infty)$
$(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$,then $f(x)$ is increasing in	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

If $A$ is a $2 \times 2$ matrix such that $\operatorname{det} A = -21$ and $\operatorname{trace}(A^3) = 2024$, then the trace of $A$ is

Let $A$ be a symmetric matrix and $B$ be a skew-symmetric matrix,such that $A - B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Then $|A|$ is:

If $A$ is a skew-symmetric matrix of order $3$ and $X$ is another matrix of the same order,then $|XA + AX^T|$ is (where $|P|$ denotes the determinant of matrix $P$).