The number of $3 \times 3$ matrices $A$, whose entries are either $1$ or $-1$ and for which the system $A\left[ {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
0
\end{array}} \right]$ has exactly three distinct solutions, is

Let $A=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right], x \in R$ and $A^{4}=\left[a_{i j}\right] .$ If $a_{11}=109,$ then $a_{22}$ is equal to

If matrix $A = \left[ {\begin{array}{*{20}{c}}0&{ – 1}\\1&0\end{array}} \right]$, then ${A^{16}} = $

If $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&{ – i}\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}0&i\\i&0\end{array}} \right]$, where $i = \sqrt { – 1} $, then the correct relation is

Let $A=\left(\begin{array}{rrr}1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{array}\right)$ and $B=7 A^{20}-20 A^{7}+2 I$, where $I$ is an identity matrix of order $3 \times 3$ If $B=\left[b_{i j}\right]$, then $b_{13}$ is equal to $….$