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

Among the statements:
$I$: If $\begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} = \begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{vmatrix}$, then $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=\frac{3}{2}$
$II$: If $\begin{vmatrix} x^{2}+x & x+1 & x-2 \\ 2x^{2}+3x-1 & 3x & 3x-3 \\ x^{2}+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = px+q$, then $p^{2}=196q^{2}$

If $A = \int\limits_1^{\sin \theta } {\frac{t}{{1 + {t^2}}}} dt$ and $B = \int\limits_1^{\csc \theta } {\frac{dt}{{t\left( {1 + {t^2}} \right)}}} $,(where $\theta \in \left( {0, \frac{\pi }{2}} \right)$),then the value of $\left| {\begin{array}{*{20}{c}} A & {{A^2}} & { - B} \\ {{e^{A + B}}} & {{B^2}} & { - 1} \\ 1 & {{A^2} + {B^2}} & { - 1} \end{array}} \right|$ is

Let $P$ be an $m \times m$ matrix such that $P^2=P$. Then,$(I+P)^n$ equals

If $ab + bc + ca = 0$ and $\begin{vmatrix} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{vmatrix} = 0$,then one of the values of $x$ is

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$, then $A^3 - 4A^2 - 6A$ is equal to: