Animated Solution for Mathematics - Matrices and Determinants: How many $3\times 3$ matrices M with entries from $\{0,1,2\}$ are there, for which the sum of the diagonal entries of $M^{T}M$ is 5?

Welcome, future engineer. Today we stand before a problem that, at first glance, appears to be a daunting exercise in linear algebra. We are asked to count the number of $3\times 3$ matrices $M$ with entries from the set $\{0,1,2\}$ such that the trace of $M^{T}M$ is exactly $5$.

Many students see the term $\text{Tr}(M^{T}M)$ and immediately reach for the definition of matrix multiplication, preparing to write out nine separate equations for the diagonal elements. But I want you to pause. Take a breath. In JEE Advanced, the most powerful weapon you possess is not brute force, but insight. Let us unmask this problem together.

First, let us demystify the expression $\text{Tr}(M^{T}M)$. If $M$ is a $3\times 3$ matrix with entries $m_{ij}$, then the $k$-th diagonal entry of the product $M^{T}M$ is the dot product of the $k$-th column of $M$ with itself.

When we sum these diagonal entries to find the trace, we are effectively summing the squares of every single entry in the matrix. Mathematically, this is expressed as:

Suddenly, the matrix structure vanishes, and we are left with a beautiful, simple constraint: the sum of the squares of the nine entries must equal $5$. This is the turning point. We have moved from the abstract world of matrices to the concrete world of counting.

We have nine slots to fill. Let $x_0$ be the number of zeros, $x_1$ be the number of ones, and $x_2$ be the number of twos. We have two fundamental constraints.

First, the total number of entries is nine:

Second, the sum of squares is five:

This is a Diophantine equation—an equation where we seek integer solutions. Because $x_1$ and $x_2$ must be non-negative, we can test values for $x_2$. If $x_2=2$, then $4(2)=8$, which is already greater than $5$. This is impossible. Thus, $x_2$ can only be $0$ or $1$.

Substituting this into our equation $x_1+4x_2=5$, we find $x_1=5$. Then, using our total count equation $x_0+x_1+x_2=9$, we get $x_0+5+0=9$, which means $x_0=4$.

We have four zeros and five ones. The number of ways to arrange these is given by the multinomial coefficient:

Substituting this into $x_1+4x_2=5$, we find $x_1=1$. Then, using $x_0+x_1+x_2=9$, we get $x_0+1+1=9$, which means $x_0=7$.

We have seven zeros, one one, and one two. The number of ways to arrange these is:

We have explored the landscape, identified the constraints, and solved the cases. All that remains is to sum our results:

You have navigated the complexity of matrix algebra and arrived at the solution through logical deduction. Remember, the next time you see a matrix trace, do not fear the multiplication. Look for the sum of squares, look for the symmetry, and trust your mathematical intuition. The final answer is 198.