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

If $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $AB = O$,then $B =$

Consider the matrices $A = \begin{bmatrix} 4 & 6 & -1 \\ 3 & 0 & 2 \\ 1 & -2 & 5 \end{bmatrix}$,$B = \begin{bmatrix} 2 & 4 \\ 0 & 1 \\ -1 & 2 \end{bmatrix}$,and $C = \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}$. Which of the following matrix products are defined?
$(i) (AB)^T C$
$(ii) C^T C (AB)^T$
$(iii) C^T AB$
$(iv) A^T AB B^T C$

If $A = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}$,then ${A^2} = $

Let $n \geq 2$ be an integer. $A = \begin{bmatrix} \cos (2 \pi / n) & \sin (2 \pi / n) & 0 \\ -\sin (2 \pi / n) & \cos (2 \pi / n) & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $I$ is the identity matrix of order $3$. Then,

If $A$ is a square matrix such that $A^2 = A$,then $(I + A)^2 - 3A =$ . . . . . . .