1. Key Concept: Trace of $A^T A$

The problem asks for the sum of all diagonal elements of the matrix product $A^T A$. This sum is formally known as the trace of the matrix $A^T A$, denoted as $\mathrm{Tr}(A^T A)$.

A fundamental and highly useful property for any real matrix $A$ is that the trace of $A^T A$ is equal to the sum of the squares of all its individual elements. Let $A = [a_{ij}]$ be an $m \times n$ matrix with real entries. Then: $\mathrm{Tr}(A^T A) = \sum_{i=1}^m \sum_{j=1}^n a_{ij}^2$ Why this is important: This property significantly simplifies the problem. Instead of performing matrix multiplication $A^T A$ and then summing its diagonal elements, we can directly work with the individual entries of matrix $A$. This avoids complex calculations and makes the problem tractable.

Proof of the Property (for clarity): Let $A$ be an $m \times n$ matrix. Let $C = A^T A$. The element $c_{ik}$ of $C$ is given by: $c_{ik} = \sum_{j=1}^n (A^T)_{ij} A_{jk}$ Since $(A^T)_{ij} = a_{ji}$, we have: $c_{ik} = \sum_{j=1}^n a_{ji} a_{jk}$ The diagonal elements of $C$ are $c_{ii}$. So, $c_{ii} = \sum_{j=1}^n a_{ji} a_{ji} = \sum_{j=1}^n (a_{ji})^2$ The trace of $C$ is the sum of its diagonal elements: $\mathrm{Tr}(A^T A) = \sum_{i=1}^m c_{ii} = \sum_{i=1}^m \sum_{j=1}^n (a_{ji})^2$ Since the sum is over all $m \times n$ elements, the order of indices $i$ and $j$ in $a_{ji}$ doesn't change the sum. Thus, $\sum_{i=1}^m \sum_{j=1}^n (a_{ji})^2 = \sum_{i=1}^m \sum_{j=1}^n a_{ij}^2$.

2. Applying the Concept to the Problem

The problem states that $A$ is a $3 \times 3$ matrix, meaning it has $3 \times 3 = 9$ elements. Let these elements be $a_{ij}$, where $i, j \in \{1, 2, 3\}$. The entries $a_{ij}$ are chosen from the set $\{-1, 0, 1\}$. We are given that the sum of all diagonal elements of $A^T A$ is 6. Using the key property: $\sum_{i=1}^3 \sum_{j=1}^3 a_{ij}^2 = 6$

3. Analyzing the Sum of Squares

Now, let's consider the possible values for $a_{ij}^2$ given that $a_{ij} \in \{-1, 0, 1\}$:

• If $a_{ij} = -1$, then $a_{ij}^2 = (-1)^2 = 1$.
• If $a_{ij} = 0$, then $a_{ij}^2 = (0)^2 = 0$.
• If $a_{ij} = 1$, then $a_{ij}^2 = (1)^2 = 1$. So, each term $a_{ij}^2$ in the sum can only be either 0 or 1.

We have a sum of 9 such terms, and their total sum must be 6: $a_{11}^2 + a_{12}^2 + \dots + a_{33}^2 = 6$ Since each $a_{ij}^2$ is either 0 or 1, for their sum to be 6, exactly 6 of these terms must be 1, and the remaining $9-6=3$ terms must be 0. Why this is important: This deduction translates the matrix problem into a combinatorial counting problem: we need to count how many ways we can choose the entries of $A$ such that 6 of them are non-zero (i.e., $\pm 1$) and 3 are zero.

4. Counting the Possibilities

To construct a matrix $A$ that satisfies the condition, we need to make two decisions:

1. Choose the positions for the non-zero elements: We need to select 6 positions out of the 9 available positions in the $3 \times 3$ matrix where the entries will be non-zero (i.e., $\pm 1$). The number of ways to do this is given by the binomial coefficient $\binom{9}{6}$. $\binom{9}{6} = \binom{9}{9-6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$
2. Assign values to the chosen positions:
   • For each of the 6 chosen positions, the entry $a_{ij}$ must be such that $a_{ij}^2 = 1$. This means $a_{ij}$ can be either $-1$ or $1$. There are 2 choices for each of these 6 positions. Since these choices are independent, the total number of ways to assign values to these 6 positions is $2^6$. $2^6 = 64$
   • For the remaining 3 positions (which were not chosen in step 1), the entry $a_{ij}$ must be such that $a_{ij}^2 = 0$. This means $a_{ij}$ must be $0$. There is only 1 choice for each of these 3 positions. The total number of ways to assign values to these 3 positions is $1^3 = 1$.

Total Number of Matrices: The total number of matrices $A$ satisfying the condition is the product of the possibilities from these steps: Total number of matrices = (Ways to choose positions for non-zero elements) $\times$ (Ways to assign values to non-zero elements) $\times$ (Ways to assign values to zero elements) $\text{Total matrices} = \binom{9}{6} \times 2^6 \times 1^3 = 84 \times 64 \times 1$

5. Calculation

$84 \times 64 = 5376$

Therefore, there are 5376 matrices $A \in S$ such that the sum of all the diagonal elements of $A^T A$ is 6.

Tips and Common Mistakes:

• Don't forget the transpose: The property $\mathrm{Tr}(A^T A) = \sum a_{ij}^2$ is crucial. A common mistake is to misinterpret "diagonal elements of $A^T A$" or to incorrectly calculate $A^T A$.
• Careful with entry values: Remember that $a_{ij}^2$ is always non-negative. For entries from $\{-1,0,1\}$, $a_{ij}^2$ can only be $0$ or $1$. This simplifies the sum of squares constraint greatly.
• Combinatorics: Problems involving counting matrices often combine matrix properties with combinatorial principles (like combinations and permutations). Ensure you correctly identify which elements are fixed, which have choices, and how many ways those choices can be made.
• Consistency: Double-check that your interpretation of the problem statement (e.g., "sum of all diagonal elements" meaning trace) is consistent with standard definitions.

Summary and Key Takeaway: The key to solving this problem efficiently is recognizing the identity $\mathrm{Tr}(A^T A) = \sum_{i,j} a_{ij}^2$. Once this property is applied, the problem transforms into a straightforward combinatorics question. For a $3 \times 3$ matrix with entries from $\{-1,0,1\}$, the condition $\mathrm{Tr}(A^T A) = 6$ implies that exactly 6 of the matrix elements must be $\pm 1$ and the remaining 3 must be $0$. The number of ways to choose these positions and assign the values leads to 5376 possible matrices.