If matrix $A = [a_{ij}]_{3 \times 3}$ and $B = [b_{ij}]_{3 \times 3}$,where $a_{ij} + a_{ji} = 0$ and $b_{ij} - b_{ji} = 0$ for all $i, j$,then $A^4B^3$ is:

Let $A$ be a matrix such that $A \cdot \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ is a scalar matrix and $|3A| = 108$. Then $A^2$ equals

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 $I+A = (I-A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.

Let $A = \begin{bmatrix} m & n \\ p & q \end{bmatrix}$,$d = |A| \neq 0$ and $|A - d(\operatorname{Adj} A)| = 0$. Then:

$\det \left[ \begin{array}{ccc} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{array} \right] = $

Let $A = (a_{ij})$ be an $n \times n$ matrix defined by $a_{ij} = \begin{cases} k^i, & \forall i=j \\ 0, & \text{otherwise} \end{cases}$. If $m = \text{trace of } A$ and $\lim_{k \rightarrow 1} \frac{n-m}{1-k} = 171$, then the value of $n$ is: