Key Concept: Trace of a Matrix and the Sum of Squares

The problem involves the sum of diagonal elements of the matrix product $AA^T$. This sum is formally known as the trace of the matrix $AA^T$, denoted as $\text{Tr}(AA^T)$. A crucial property for any $n \times n$ matrix $A$ with entries $a_{ij}$ is that: $\text{Tr}(AA^T) = \sum_{i=1}^n \sum_{j=1}^n a_{ij}^2$ This formula states that the sum of the diagonal elements of $AA^T$ is equal to the sum of the squares of all entries of the matrix $A$. This property is fundamental to solving this problem.

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Step-by-Step Derivation

1. Define the Matrix A and its Transpose $A^T$ Let $A$ be a $3 \times 3$ matrix with entries $a_{ij}$ from the set $\{-1, 0, 1\}$. $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$ The transpose of $A$, denoted $A^T$, is obtained by interchanging rows and columns: $A^T = \begin{pmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{pmatrix}$

2. Calculate the Diagonal Elements of $AA^T$ To find the diagonal elements of $AA^T$, we multiply the $i$-th row of $A$ by the $i$-th column of $A^T$. Let $(AA^T)_{ii}$ denote the $i$-th diagonal element.

For the first diagonal element $(AA^T)_{11}$: It is the dot product of the first row of $A$ and the first column of $A^T$. $(AA^T)_{11} = (a_{11}, a_{12}, a_{13}) \cdot (a_{11}, a_{12}, a_{13}) = a_{11}^2 + a_{12}^2 + a_{13}^2$ For the second diagonal element $(AA^T)_{22}$: It is the dot product of the second row of $A$ and the second column of $A^T$. $(AA^T)_{22} = (a_{21}, a_{22}, a_{23}) \cdot (a_{21}, a_{22}, a_{23}) = a_{21}^2 + a_{22}^2 + a_{23}^2$ For the third diagonal element $(AA^T)_{33}$: It is the dot product of the third row of $A$ and the third column of $A^T$. $(AA^T)_{33} = (a_{31}, a_{32}, a_{33}) \cdot (a_{31}, a_{32}, a_{33}) = a_{31}^2 + a_{32}^2 + a_{33}^2$

3. Formulate the Given Condition The problem states that the sum of the diagonal elements of $AA^T$ is 3. Using the calculations from Step 2, this sum is: $(a_{11}^2 + a_{12}^2 + a_{13}^2) + (a_{21}^2 + a_{22}^2 + a_{23}^2) + (a_{31}^2 + a_{32}^2 + a_{33}^2) = 3$ This can be written more compactly as the sum of squares of all 9 entries of matrix $A$: $\sum_{i=1}^3 \sum_{j=1}^3 a_{ij}^2 = 3$

4. Analyze the Properties of the Entries The entries $a_{ij}$ can only take values from the set $\{-1, 0, 1\}$. Let's consider the possible values for $a_{ij}^2$:

• 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 $a_{ij}^2$ can only be either $0$ or $1$.

Now, we have the sum of nine such terms ($a_{ij}^2$) equal to 3: $a_{11}^2 + a_{12}^2 + \dots + a_{33}^2 = 3$ Since each $a_{ij}^2$ is either 0 or 1, for their sum to be exactly 3, it must be the case that exactly three of these $a_{ij}^2$ terms are equal to 1, and the remaining $9-3=6$ terms are equal to 0.

• If $a_{ij}^2 = 1$, then $a_{ij}$ can be either $1$ or $-1$. (2 choices)
• If $a_{ij}^2 = 0$, then $a_{ij}$ must be $0$. (1 choice)

5. Count the Number of Possible Matrices We need to construct matrices $A$ that satisfy these conditions. This is a combinatorial problem involving two stages:

• Stage 1: Choose the positions for the non-zero entries. There are 9 total entries in the $3 \times 3$ matrix. We need to select 3 of these positions to have non-zero values (i.e., $a_{ij}^2 = 1$). The number of ways to choose these 3 positions out of 9 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$: $\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$ So, there are 84 ways to choose which three entries in the matrix will be non-zero.

• Stage 2: Assign values to the chosen non-zero entries. For each of the 3 chosen positions, the entry $a_{ij}$ can be either 1 or -1. Since there are 2 choices for each of these 3 positions, the total number of ways to assign values to these non-zero entries is $2 \times 2 \times 2 = 2^3 = 8$.

• Stage 3: Assign values to the remaining zero entries. The remaining $9-3=6$ entries must have $a_{ij}^2 = 0$, which means $a_{ij}$ must be 0. There is only 1 way to assign 0 to each of these positions. This step doesn't multiply the total count further as $1^6=1$.

Total Number of Matrices: To find the total number of such matrices, we multiply the number of ways from Stage 1 and Stage 2: $\text{Number of matrices} = \binom{9}{3} \times 2^3 = 84 \times 8 = 672$

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Tips and Common Mistakes:

1. Understanding the Trace: Ensure you understand the property $\text{Tr}(AA^T) = \sum a_{ij}^2$. This property significantly simplifies the problem by converting a matrix multiplication problem into a sum of squares problem.
2. Entry Value Analysis: Carefully analyze the possible values of $a_{ij}$ and $a_{ij}^2$. For entries from $\{-1, 0, 1\}$, $a_{ij}^2$ can only be 0 or 1. This is the critical step to deduce how many entries must be non-zero.
3. Combinatorics vs. Permutations: This problem involves choosing positions and then assigning values. Since the positions are distinct (e.g., $a_{11}$ is different from $a_{12}$), we use combinations to choose the locations of the non-zero entries, and then for each chosen location, we have independent choices for its value.
4. Don't forget the sign choices: A common mistake is to only count the ways to choose positions ($\binom{9}{3}$) and forget that the non-zero entries can be either 1 or -1, leading to an answer of 84. The $2^3$ factor is crucial.

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Summary and Key Takeaway:

This problem effectively tests your understanding of matrix operations (specifically $AA^T$), the concept of a matrix trace, and basic combinatorics. The key insight is realizing that the condition on the sum of diagonal elements translates to the sum of squares of all entries of the matrix. Given the restricted set of allowed entries $\{-1, 0, 1\}$, this sum of squares directly dictates how many entries must be non-zero and what values they can take. The solution then becomes a straightforward application of combinations to choose positions and powers to assign values.

The final answer is $\boxed{672}$.