Let $S = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{0, 1, 2, 3, 4 \} \text{ and } A^2 - 4A + 3I = 0 \right\}$ be a set of $2 \times 2$ matrices. Then the number of matrices in $S$, for which the sum of the diagonal elements is equal to $4$, is:

If $A$ and $B$ are two invertible square matrices of the same order such that $(A + B)(A - B) = A^2 - B^2$,then $(A^2BA^{-1}B^{-1})^3$ is equal to-

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{T} = \alpha A + I$,where $\alpha \in R - \{-1, 1\}$. If $\det(A^2 - A) = 4$,then the sum of all possible values of $\alpha$ 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

If $\left| \begin{array}{ccc} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{array} \right| = 0$ and the vectors $\vec{a} = (1, a, a^2)$,$\vec{b} = (1, b, b^2)$,and $\vec{c} = (1, c, c^2)$ are non-coplanar,then $abc$ is equal to

If $A=\begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix}$,$B=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$,$C=ABA^T$ and $X=A^T C^2 A$,then $\operatorname{det}(X)$ is equal to: