Let $A$ and $B$ be two $3 \times 3$ non-singular matrices such that $\operatorname{det}(A^T B A) = 27$ and $\operatorname{det}(A B^{-1}) = 8$. Then $\operatorname{det}(B^T A^{-1} B) = $

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 $\theta \in \left( {0, \frac{\pi }{2}} \right)$:

Let $S =\{ M = [a_{ij}], a_{ij} \in \{0,1,2\}, 1 \leq i, j \leq 2\}$ be a sample space and $A = \{M \in S : M \text{ is invertible}\}$ be an event. Then $P(A)$ is equal to

If $A$ is a square matrix of $3 \times 3$ order,and $|A| = 2$,then $|(A-A^T)^6| + |(A^T-A)^7|$ is equal to (where $A^T$ denotes the transpose of matrix $A$).

$A=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \Rightarrow A^2-2A=$

Let $A = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$, and $P = \begin{bmatrix} 0 & 1 & 0 \\ x & 0 & 0 \\ 0 & 0 & y \end{bmatrix}$ be an orthogonal matrix such that $B = PAP^{-1}$ holds. Then: