Let A = [ cases 0 & - 2 2 & 0 cases ] . If M and N are two matrices...

Since A = \left[\begin{cases}0 & - 2 \\ 2 & 0\end{cases}\right]

So {A}^{2} = \left[\begin{cases}0 & - 2 \\ 2 & 0\end{cases}\right]\left[\begin{cases}0 & - 2 \\ 2 & 0\end{cases}\right] = \left[\begin{cases}- 4 & 0 \\ 0 & - 4\end{cases}\right] = - 4I

and ${A}^{3} = - 4A$

Similarly ${A}^{4} = \left(- 4I\right)\left(- 4I\right) = {\left(- 4\right)}^{2}I$,

${A}^{5} = {\left(- 4\right)}^{2}A, {A}^{6} = {\left(- 4\right)}^{3}I$

Now $M = \mathop{∑}_{k = 1}^{10}{A}^{2k} = {A}^{2} + {A}^{4} + \ldots . + {A}^{20}$

$= \left[- 4 + {\left(- 4\right)}^{2} + {\left(- 4\right)}^{3} + \ldots . + {\left(- 4\right)}^{20}\right]I$

$= - {k}_{1}I$

So $M$ is symmetric matrix

$N = \mathop{∑}_{k = 1}^{10}{A}^{2k - 1} = A + {A}^{3} + \ldots \ldots + {A}^{19}$

$= A\left[1 + \left(- 4\right) + {\left(- 4\right)}^{2} + \ldots \ldots + {\left(- 4\right)}^{9}\right]$

$= {k}_{2}A$

So $N$ is skew symmetric

$⇒ {N}^{2}$ is symmetric matrix

Hence, $M{N}^{2}$ is non-identity symmetric matrix