Find $X$ and $Y$,if $X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$ and $X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$.

The matrix $A = \frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ -2 & 2 & -1 \end{bmatrix}$ is

Assume $X, Y, Z, W$ and $P$ are the matrices of order $2 \times n, 3 \times k, 2 \times p, n \times 3$ and $p \times k$ respectively. The restriction on $n, k$ and $p$ so that $PY + WY$ will be defined are:

If $A^{\prime}=\begin{bmatrix}-2 & 3 \\ 1 & 2\end{bmatrix}$ and $B=\begin{bmatrix}-1 & 0 \\ 1 & 2\end{bmatrix},$ then find $(A+2B)^{\prime}$.

If $A$ and $B$ are $n \times n$ square matrices such that $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$,then $A B A B=$

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