Animated Solution for Mathematics - Matrices and Determinants: The number of all $3\times 3$ matrices $A$, with entries from the set $\{-1,0,1\}$ such that the sum of the diagonal elements of $A{A}^T$ is $3$, is .....

Welcome, fellow traveler on the road to JEE Advanced. Today, we are not just solving a matrix problem; we are uncovering a hidden structure within a $3\times 3$ grid.

When you see a matrix $A$ with entries from $\{-1,0,1\}$, it is easy to feel overwhelmed by the sheer number of possibilities. But remember, in mathematics, constraints are not barriers—they are clues. Let us peel back the layers of this problem together.

We are given the condition $Tr(A{A}^T)=3$. At first glance, this looks like a standard linear algebra identity.

If you perform the matrix multiplication, you will find that the diagonal elements of $A{A}^T$ are the sums of the squares of the elements of the rows of $A$. Specifically, the $(i,i)$-th entry of $A{A}^T$ is ${\sum }_{j=1}^3{a}_{ij}^2$.

When we take the trace, we sum these diagonal elements, which effectively gives us the sum of the squares of every single entry in the matrix:

This is the "Aha!" moment. We have 9 slots in our $3\times 3$ matrix, and the sum of the squares of the values in these slots must be exactly 3. Since our entries $a_{ij}$ are restricted to $\{-1,0,1\}$, their squares $a_{ij}^2$ can only be $0$ or $1$.

If we have 9 slots and each slot contributes either $0$ or $1$ to the total sum, and that total sum must be 3, then exactly 3 slots must contain a non-zero value, and the remaining 6 slots must be zero. This transforms our matrix problem into a classic selection problem.

First, we must choose which 3 positions out of the 9 will hold our non-zero values. The number of ways to choose these positions is given by the binomial coefficient $\left(\genfrac{}{}{0ex}{}{9}{3}\right)$:

So, there are 84 different "skeletons" of matrices where exactly three entries are non-zero. But we are not done yet, as each of those 3 non-zero entries can be either $1$ or $-1$.

For each of the 3 chosen positions, we have 2 choices. Since these choices are independent, for every one of the 84 skeletons, there are $2\times 2\times 2=2^3=8$ ways to populate the non-zero entries.

To find the total number of such matrices, we simply multiply the number of ways to choose the positions by the number of ways to assign the values:

There it is. 672. It is a beautiful, clean result that emerges from a seemingly chaotic set of possibilities.

Whenever you face a problem that seems to involve "all possible matrices," stop and look for the invariant—the property that remains constant, like our sum of squares. Once you find that, the complexity melts away, leaving only the elegant logic of counting. Keep practicing, keep questioning, and most importantly, keep finding the beauty in the math.