Let A=pmatrix0&-22&0pmatrix.

Question

Let $A=\begin{pmatrix}0&-2\2&0\end{pmatrix}$. If $M=\displaystyle\sum_{k=1}^{10}A^{2k}$ and $N=\displaystyle\sum_{k=1}^{10}A^{2k-1}$, then $MN^2$ is:

Accepted Answer

$A^2=\begin{pmatrix}-4&0\0&-4\end{pmatrix}=-4I$, so $A^{2k}=(-4)^kI$ and $A^{2k-1}=(-4)^{k-1}A$.

$M=\displaystyle\sum_{k=1}^{10}(-4)^kI$ is a scalar multiple of $I$ (hence symmetric), and $N=\displaystyle\sum_{k=1}^{10}(-4)^{k-1}A$ is a scalar multiple of $A$ (hence, since $A$ is skew-symmetric, $N$ is skew-symmetric, and $N^2$ is a (negative) scalar multiple of $I$, hence symmetric).

So $MN^2$, being a product of two scalar multiples of $I$-and-$A^2$-type matrices, is itself a scalar multiple of $I$: a non-identity symmetric matrix.

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