Mathematics - Matrices Question with Solution | TestHub

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

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

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 $M=\sum _{k=1}^{10}{A}^{2k}=A^2+A^4+\dots .+A^{20}$

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

$=-k_{1}I$

So $M$ is symmetric matrix

$N=\sum _{k=1}^{10}{ A}^{2k-1}=A+A^3+\dots \dots +A^{19}$

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

$=k_{2}A$

So $N$ is skew symmetric

$\Rightarrow N^2$ is symmetric matrix

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