If $A = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$,then the determinant of $A^2 - 2A$ is

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:

$A$ and $B$ are two non-singular square matrices of order $3 \times 3$ such that $AB = A$ and $|A + B| \neq 0$. Then:

Consider the matrix $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Given below are two statements:
Statement $I$: $f(-x)$ is the inverse of the matrix $f(x)$.
Statement $II$: $f(x) f(y) = f(x+y)$.
In the light of the above statements,choose the correct answer from the options given below:

Let $A = \begin{bmatrix} 5 & \sin^2 \theta & \cos^2 \theta \\ -\sin^2 \theta & -5 & 1 \\ \cos^2 \theta & 1 & 5 \end{bmatrix}$. Then the maximum value of $\det(A)$ is

Let $B=\begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}, \alpha > 2$ be the adjoint of a matrix $A$ and $|A|=2$. Then the value of $\begin{bmatrix} \alpha & -2\alpha & \alpha \end{bmatrix} B \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix}$ is equal to: