An $n \times n$ matrix is formed using $0, 1$ and $-1$ as its elements. The number of such matrices which are skew-symmetric is

If $A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}$,then $A^n = 2^k A$,where $k = $

If $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$,then $AB = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ and $BA = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Clearly $AB \neq BA$. Thus,matrix multiplication is not commutative. Is this statement true for all matrices?

Let $A$ be a square matrix such that $a_{ij} \in \{-1, 0, 1\}$ for all $i, j$ and it has exactly one non-zero entry in each row as well as in each column. Then:

If $A = \begin{bmatrix} 0 & 2 & 0 \\ 0 & 0 & 3 \\ -2 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & -4 & 0 \end{bmatrix}$,then the element of the $3^{rd}$ row and $3^{rd}$ column in $AB$ will be:

If $\begin{bmatrix} x & 0 \\ 1 & y \end{bmatrix} + \begin{bmatrix} -2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 6 & 3 \end{bmatrix} - \begin{bmatrix} 2 & 4 \\ 2 & 1 \end{bmatrix}$,then find the values of $x$ and $y$.