Let A=[ … ]. If M and N are two matrices given by M=Σ _k=1^10 A^2k and N=Σ _k=1^10 A^2k-1 then MN^2 is

Since $A=[\begin{matrix}0 & -2 \\ 2 & 0\end{matrix}]$A=[02​−20​]

So ${A}^{2}=[\begin{matrix}0 & -2 \\ 2 & 0\end{matrix}][\begin{matrix}0 & -2 \\ 2 & 0\end{matrix}]=[\begin{matrix}-4 & 0 \\ 0 & -4\end{matrix}]=-4I$A2=[02​−20​][02​−20​]=[−40​0−4​]=−4I

and ${A}^{3}=-4A$A3=−4A

Similarly ${A}^{4}=(-4I)(-4I)={(-4)}^{2}I$A4=(−4I)(−4I)=(−4)2I,

$${A}^{5}={(-4)}^{2}A,{A}^{6}={(-4)}^{3}I$$A5=(−4)2A,A6=(−4)3I

Now $\displaystyle M=\sum _{k=1}^{10}{A}^{2k}={A}^{2}+{A}^{4}+\ldots .+{A}^{20}$M=k=1∑10​A2k=A2+A4+….+A20

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

$$=-{k}_{1}I$$=−k1​I

So $M$M is symmetric matrix

$$N=\sum _{k=1}^{10}{A}^{2k-1}=A+{A}^{3}+\ldots \ldots +{A}^{19}$$N=k=1∑10​A2k−1=A+A3+……+A19

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

$$={k}_{2}A$$=k2​A

So $N$N is skew symmetric

$\Rightarrow {N}^{2}$⇒N2 is symmetric matrix

Hence, $M{N}^{2}$MN2 is non-identity symmetric matrix