Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]$. 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 $M{N}^2$ is

Welcome, fellow traveler in the realm of linear algebra. Today, we are going to dismantle a problem that, at first glance, looks like a tedious exercise in matrix exponentiation.

You see a sum of powers of $A$ up to $A^{20}$, and your instinct might be to panic. But wait—take a breath. In JEE Advanced, the complexity is often a mask for a beautiful, underlying simplicity. Let us peel back that mask together.

We are given the matrix $A=\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]$. Whenever you see a matrix with zeros on the diagonal and anti-symmetric off-diagonal elements, your intuition should scream 'rotation' or 'cyclic behavior.'

Let us test this by calculating the square of $A$:

Do you see it? $A^2$ is not just any matrix; it is a scalar multiple of the identity matrix $I$. This is our 'Golden Key.'

Because $A^2=-4I$, any higher power of $A$ will collapse into a simple scalar multiple of either $I$ or $A$. This observation transforms a nightmare of matrix multiplication into a simple geometric series problem.

Now, let us look at $M={\sum }_{k=1}^{10}{A}^{2k}$. Since $A^{2k}=(A^2{)}^k=(-4I{)}^k=(-4{)}^{k}I$, we can factor out the identity matrix:

Let $\lambda ={\sum }_{k=1}^{10}(-4{)}^k$. This is just a number—a scalar. So, $M=\lambda I$.

Similarly, for $N={\sum }_{k=1}^{10}{A}^{2k-1}$, we can factor out one $A$:

Let $\mu ={\sum }_{j=0}^9(-4{)}^j$. Now we have $N=\mu A$. The entire complexity of the original matrices has been reduced to two simple scalar-matrix products.

We are tasked with finding the nature of $M{N}^2$. Substituting our simplified forms:

Since $I$ is the identity, $I\cdot A^2=A^2$. And we know $A^2=-4I$. Thus:

Let $k=-4\lambda {\mu }^2$. We have arrived at $M{N}^2=kI$.

What is the nature of $kI$? By definition, a matrix is symmetric if it equals its own transpose.

Since $(kI{)}^T=k{I}^T=kI$, any scalar multiple of the identity matrix is symmetric. Because $k$ is clearly not $1$, we have successfully proven that $M{N}^2$ is a non-identity symmetric matrix.

Isn't it elegant? We started with a daunting sum of matrix powers and ended with the realization that the entire expression is just a scaled version of the identity. Keep this perspective in your toolkit: whenever you see high powers of a matrix, look for a pattern that reduces it to the identity. You have mastered this concept today!