Let $$A = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$$. If $$M$$ and $$N$$ are two matrices given by $$M = \sum_{k=1}^{10} A^{2k}$$ and $$N = \sum_{k=1}^{10} A^{2k-1}$$ then $$MN^2$$ is

We are given $$A = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$$, and we wish to determine the nature of $$MN^2$$ where $$M = \sum_{k=1}^{10} A^{2k}$$ and $$N = \sum_{k=1}^{10} A^{2k-1}$$.

Since $$A^2 = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} -4 & 0 \\ 0 & -4 \end{pmatrix} = -4I$$, it follows that $$A^{2k} = (-4)^k I$$ for any integer $$k\ge1$$.

Substituting this into the definition of $$M$$ gives $$M = \sum_{k=1}^{10} (-4)^k I = \left[\sum_{k=1}^{10}(-4)^k\right]I$$. Let $$m = \sum_{k=1}^{10}(-4)^k = \frac{-4((-4)^{10} - 1)}{-4 - 1} = \frac{-4(4^{10} - 1)}{-5} = \frac{4(4^{10} - 1)}{5}$$, so that $$M = mI$$ where $$m = \frac{4(4^{10}-1)}{5}$$.

Similarly, since $$A^{2k-1} = A^{2(k-1)}A = (-4)^{k-1}A$$, one finds $$N = \sum_{k=1}^{10}(-4)^{k-1}A = \left[\sum_{k=0}^{9}(-4)^k\right]A$$. Defining $$n = \sum_{k=0}^{9}(-4)^k = \frac{1-(-4)^{10}}{1-(-4)} = \frac{1-4^{10}}{5} = -\frac{4^{10}-1}{5}$$ yields $$N = nA$$.

Proceeding to $$N^2$$, we have $$N^2 = n^2A^2 = n^2(-4I) = -4n^2I$$. Consequently, $$MN^2 = (mI)(-4n^2I) = -4mn^2I$$, which is a scalar multiple of the identity matrix.

Using $$m = \frac{4(4^{10}-1)}{5}$$ and $$n^2 = \frac{(4^{10}-1)^2}{25}$$, it follows that $$-4mn^2 = -4\cdot\frac{4(4^{10}-1)}{5}\cdot\frac{(4^{10}-1)^2}{25} = \frac{-16(4^{10}-1)^3}{125}$$, which is clearly not equal to 1. Therefore, $$MN^2$$ is a non-identity scalar matrix. Since any scalar matrix $$cI$$ satisfies $$(cI)^T = cI$$, it is symmetric.

Hence, the correct classification is that $$MN^2$$ is a non-identity symmetric matrix, corresponding to Option A.