If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$,then $A^{2} - 5A$ is equal to:

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 $A = [a\, b]$,$B = [-b\, -a]$ and $C = \begin{bmatrix} a \\ -a \end{bmatrix}$,then the correct statement is

$A$ square matrix $[a_{ij}]_{n \times n}$ is an upper triangular matrix if:

If $A = \begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$,then $A^2 - 6A =$ . . . . . . (in $I_3$)

The number of matrices of order $3 \times 3$,whose entries are either $0$ or $1$ and the sum of all the entries is a prime number,is: