Animated Solution for Mathematics - Matrices and Determinants: The total number of $3\times 3$ matrices $A$ having entries from the set $\{0,1,2,3\}$ such that the sum of all the diagonal entries of $A{A}^T$ is 9, is equal to ______

Imagine you are standing before a $3\times 3$ grid. You have nine empty cells, and your toolkit contains only four numbers: $\{0,1,2,3\}$.

Your mission is to fill these cells such that the trace of the product $A{A}^T$ is exactly $9$. At first glance, this looks like a daunting problem of matrix algebra. But let us peel back the layers of this mystery together.

When we talk about $\text{Tr}(A{A}^T)$, we are not just performing abstract matrix multiplication. Let us look at the structure of $A$. If $A$ is our $3\times 3$ matrix, then $A^T$ is its transpose.

When you compute the diagonal elements of the product $A{A}^T$, you are essentially calculating the dot product of each row of $A$ with itself. Mathematically, the diagonal element at position $(i,i)$ is ${\sum }_{j=1}^3{a}_{ij}^2$.

Therefore, the trace is simply the sum of the squares of every single entry in the matrix:

Suddenly, the matrix algebra vanishes, and we are left with a beautiful combinatorial puzzle. We have $9$ variables, each chosen from $\{0,1,2,3\}$, and their squares must sum to $9$. Since the squares of our allowed entries are $\{0,1,4,9\}$, we are looking for ways to partition the number $9$ into nine parts using only these four values.

To ensure we do not miss a single possibility, we must be systematic. We are looking for combinations of nine numbers from $\{0,1,4,9\}$ that sum to $9$.

Case 1: The Power of One. If we use the number $9$, we must fill the remaining $8$ slots with $0$. This gives us the set $\{9,0,0,0,0,0,0,0,0\}$.

The number of ways to arrange these is a simple permutation of a multiset:

Case 2: The Uniformity of Unity. What if we do not use $9$ or $4$? We are left with $1$s and $0$s. To get a sum of $9$ using only $1$s, we must use nine $1$s.

The set is $\{1,1,1,1,1,1,1,1,1\}$. There is only:

Case 3: The Balance of Fours. Now, let us introduce the $4$. If we use two $4$s, we have $4+4=8$. We need one more to reach $9$, so we add a $1$. The remaining six slots must be $0$.

Our set is $\{4,4,1,0,0,0,0,0,0\}$. The number of arrangements is:

Case 4: The Final Combination. What if we use only one $4$? We need $5$ more to reach $9$, so we use five $1$s. The remaining three slots are $0$.

Our set is $\{4,1,1,1,1,1,0,0,0\}$. The number of arrangements is:

We have explored every possible combination of squares that could sum to $9$. Because these cases are mutually exclusive, we simply add them up:

Look at what we have achieved! We took a complex-looking matrix problem, stripped away the intimidating notation, and reduced it to a fundamental counting exercise.

This is the heart of JEE Advanced mathematics: finding the simple, elegant truth hidden beneath the surface of a complex problem. You have mastered the logic of the system, and the final answer is 766.