Let $X = \left\{\begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \mathbb{R} \right\}$. Define $f: X \rightarrow \mathbb{R}$ by $f(A) = \operatorname{det}(A), \forall A \in X$. Then, $f$ is

Let $x \in R$ and let $P = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix}$,$Q = \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix}$ and $R = PQP^{-1}$. Then which of the following options is/are correct?
$(1)$ For $x = 1$,there exists a unit vector $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ for which $R \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$.
$(2)$ There exists a real number $x$ such that $PQ = QP$.
$(3)$ $\det R = \det \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{bmatrix} + 8$,for all $x \in R$.
$(4)$ For $x = 0$,if $R \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = 6 \begin{bmatrix} 1 \\ a \\ b \end{bmatrix}$,then $a + b = 5$.

For some $\alpha, \beta \in R$, let $A = \begin{bmatrix} \alpha & 2 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 \\ 1 & \beta \end{bmatrix}$ be such that $A^{2} - 4A + 2I = B^{2} - 3B + I = O$. Then $(\text{det}(\text{adj}(A^{3} - B^{3})))^{2}$ is equal to ....

If $S = \{x \in [0, 2\pi] : \begin{vmatrix} 0 & \cos x & -\sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \end{vmatrix} = 0\}$,then $\sum_{x \in S} \tan \left( \frac{\pi}{3} + x \right)$ is equal to

If $ A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \tan^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix} $ and $ B = \begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & -\tan^{-1}(\pi x) \end{bmatrix} $,then $ A - B $ is

Let $M$ and $N$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $MN = NM$. If $P^T$ denotes the transpose of $P$,then $M^2 N^2 (M^T N)^{-1} (M N^{-1})^T$ is equal to