Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of $A^2$ is 1, then the possible number of such matrices is:

A Matrix Journey

Welcome, fellow traveler in the world of JEE mathematics! Today, we are going to peel back the layers of a seemingly simple matrix problem.

It is easy to look at a matrix and see just a grid of numbers, but in the realm of JEE Advanced, a matrix is a living, breathing object with geometric and algebraic properties waiting to be unlocked. Let us dive into the heart of this problem.

The Anatomy of Symmetry

We begin with a symmetric matrix $A$ of order 2. What does symmetry mean for us? It means that the matrix is equal to its transpose, $A^T=A$.

If we define our matrix as $A=\left[\begin{array}{cc}a& b\\ b& c\end{array}\right]$, the symmetry condition forces the off-diagonal elements to be identical. Here, $a,b,$ and $c$ are integers.

This is our playground. We are not dealing with arbitrary real numbers; we are constrained by the rigid, discrete world of integers. This is a massive hint!

The Trace and the Power of $A^2$

The problem asks us to consider the sum of the diagonal elements of $A^2$. In linear algebra, the sum of the diagonal elements is known as the Trace.

So, we need to find $\text{Trace}(A^2)$. First, let us compute $A^2$:

Now, the trace is simply the sum of the diagonal elements: $(a^2+b^2)+(b^2+c^2)$. We are told this sum equals 1. Thus, we arrive at our master equation:

The Integer Trap

This is where the magic happens. We have an equation with three variables, but they are not just any variables—they are integers.

This means $a^2,b^2,$ and $c^2$ must be non-negative integers ($0,1,4,9,\dots$). Look at the term $2b^2$.

If $b$ were anything other than 0, say $b=1$ or $b=-1$, then $b^2=1$, and $2b^2=2$. But our equation says the total sum is 1!

Since $a^2$ and $c^2$ cannot be negative, it is physically impossible for $b$ to be non-zero. Therefore, $b$ must be 0. This is the 'Aha!' moment that collapses the complexity of the problem.

The Geometric Insight

With $b=0$, our equation simplifies beautifully to $a^2+c^2=1$. If you have ever studied coordinate geometry, this equation should make your heart skip a beat.

It is the equation of a unit circle in the $ac$-plane! We are looking for integer points $(a,c)$ that lie on this circle.

Imagine the circle centered at $(0,0)$ with radius 1. Where does it cross the integer grid? It crosses at $(1,0),(-1,0),(0,1),$ and $(0,-1)$.

These are the only four points where both coordinates are integers.

The Final Count

Since $b$ is fixed at 0, each of these four pairs $(a,c)$ defines a unique matrix $A$:

1. If $(a,c)=(1,0)$, $A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$

2. If $(a,c)=(-1,0)$, $A=\left[\begin{array}{cc}-1& 0\\ 0& 0\end{array}\right]$

3. If $(a,c)=(0,1)$, $A=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$

4. If $(a,c)=(0,-1)$, $A=\left[\begin{array}{cc}0& 0\\ 0& -1\end{array}\right]$

There we have it! There are 4 distinct matrices.

The beauty of this problem lies in how the algebraic constraints of integers and the geometric nature of the circle converge to a simple, elegant solution. Keep practicing this kind of visualization, and you will find that even the most intimidating JEE problems start to feel like old friends.