If $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix}$,then the correct relation is

Construct a $2 \times 2$ matrix,$A = [a_{ij}]$,whose elements are given by $a_{ij} = \frac{i}{j}$.

If $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$,then $A^2 = $

Let $M = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$ and $I$ be the identity matrix of order $3$. Then $M^2 - 4M =$

Choose the correct option for the matrices given below:
$\begin{aligned} & A=\left[\begin{array}{ccc}\cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \\ 0 & 0 & 1\end{array}\right] \\ & B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \\ 0 & -\sin \frac{\pi}{3} & \cos \frac{\pi}{3}\end{array}\right] \\ & C=\left[\begin{array}{ccc}\cos \frac{\pi}{6} & 0 & \sin \frac{\pi}{6} \\ 0 & 1 & 0 \\ -\sin \frac{\pi}{6} & \cos \frac{\pi}{6} & 0\end{array}\right] \\ & D=\left[\begin{array}{ccc}\cos \frac{\pi}{2} & \sin \frac{\pi}{2} & 0 \\ -\sin \frac{\pi}{2} & \cos \frac{\pi}{2} & 0 \\ 0 & 0 & 1\end{array}\right]\end{aligned}$

If $x\begin{bmatrix} 3 \\ 2 \end{bmatrix} + y\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 15 \\ 5 \end{bmatrix}$,then the values of $x$ and $y$ are: