Animated Solution for Mathematics - Matrices and Determinants: Let $S$ be the set containing all $3\times 3$ matrices with entries from $\{-1,0,1\}$. The total number of matrices $A\in S$ such that the sum of all the diagonal elements of $A^{T}A$ is 6 is

Welcome, fellow traveler on the journey of JEE Advanced mathematics. Today, we are going to unravel a problem that might look like a daunting matrix algebra task, but is actually a beautiful exercise in combinatorial logic.

We are dealing with a set $S$ of $3\times 3$ matrices where every entry is restricted to the set $\{-1,0,1\}$. Our goal is to find how many of these matrices satisfy the condition that the sum of the diagonal elements of $A^{T}A$ is exactly $6$.

Let us begin by demystifying the expression $\text{Tr}(A^{T}A)$. The trace of a matrix is the sum of its diagonal elements.

When we compute $A^{T}A$, the diagonal element at position $(i,i)$ is the dot product of the $i$-th column of $A$ with itself. If we denote the matrix $A$ as $[a_{ij}]$, then the diagonal element at $(i,i)$ is ${\sum }_{j=1}^3{a}_{ji}^2$.

When we sum these diagonal elements to find the trace, we are effectively summing the squares of every single element in the matrix. Thus, the condition $\text{Tr}(A^{T}A)=6$ is equivalent to the elegant equation:

This is our North Star. We are looking for matrices where the sum of the squares of all nine entries is $6$.

Now, consider the constraints on our entries $a_{ij}\in \{-1,0,1\}$. When we square these values, the result is quite simple: $(-1{)}^2=1$, $(1{)}^2=1$, and $(0{)}^2=0$.

This means that every squared term $a_{ij}^2$ is strictly either $0$ or $1$. We have nine such terms, and their sum must be $6$.

This is a classic combinatorial constraint! To get a sum of $6$ from nine terms that are either $0$ or $1$, exactly $6$ of these terms must be $1$, and the remaining $3$ terms must be $0$.

Now, imagine you are standing before an empty $3\times 3$ grid. You have $9$ available slots.

You need to choose $6$ of these slots to hold non-zero values (where the square is $1$) and $3$ slots to hold the value $0$. The number of ways to choose these positions is given by the binomial coefficient $\left(\genfrac{}{}{0ex}{}{9}{6}\right)$, which is the same as $\left(\genfrac{}{}{0ex}{}{9}{3}\right)$.

Let us calculate this:

So, there are $84$ different patterns of zeros and non-zeros in our matrix.

We are almost there. We have chosen the $6$ positions that will be non-zero. But wait—what are the actual values in those positions?

For each of these $6$ positions, the entry $a_{ij}$ can be either $1$ or $-1$. Since $1^2=1$ and $(-1{)}^2=1$, both choices satisfy our condition.

This means for each of the $6$ non-zero positions, we have $2$ independent choices. Therefore, the total number of ways to assign values to these $6$ positions is $2^6=64$.

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. It is the product of our combinatorial choices and our sign choices:

Performing this multiplication, we get $5376$.

And there you have it! By breaking down the matrix condition into a sum of squares and then applying combinatorial principles, we have navigated through the complexity to find the answer. Keep practicing this way of thinking—looking past the matrix notation to the underlying counting problem—and you will find that even the most intimidating JEE problems become clear and manageable.