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-

The number of matrices $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,where $a, b, c, d \in \{-1, 0, 1, 2, 3, \ldots, 10\}$,such that $A=A^{-1}$,is

Let $A = [a_{ij}]$ be a square matrix of order $2$ with entries either $0$ or $1$. Let $E$ be the event that $A$ is an invertible matrix. Then the probability $P(E)$ is:

Let $A = \begin{bmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \text{adj}(A)$. If $\det(B) = 66$, then $\det(\text{adj}(A))$ equals:

If $\left[\begin{array}{ccc}0 & 2 & a \\ b & 0 & 4 \\ -3 & c & 0\end{array}\right]$ is a skew-symmetric matrix, then $\left[\begin{array}{cc}a & b \\ b & a\end{array}\right]\left[\begin{array}{cc}b & c \\ c & b\end{array}\right]=$

If $A = \begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is the identity matrix of order $2$,show that $I+A = (I-A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.