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

For any real matrix $$A$$, the transpose-product $$A^{T}A$$ is always square and its trace equals the sum of the squares of all entries of $$A$$.

Formally, for a $$3 \times 3$$ matrix $$A=[a_{ij}]$$,

$$\operatorname{trace}(A^{T}A)=\sum_{k=1}^{3}(A^{T}A)_{kk} =\sum_{k=1}^{3}\sum_{i=1}^{3}a_{ik}^{2} =\sum_{i=1}^{3}\sum_{k=1}^{3}a_{ik}^{2}$$

Thus, the given condition “the sum of the diagonal elements of $$A^{T}A$$ is $$6$$” is equivalent to

$$\sum_{i=1}^{3}\sum_{k=1}^{3}a_{ik}^{2}=6$$

The entries of $$A$$ are taken from $$\{-1,\,0,\,1\}$$, so each squared entry is either $$0$$ or $$1$$. Therefore the total number of non-zero (±1) entries must be exactly $$6$$, because each contributes $$1$$ to the sum and we need the sum to be $$6$$.

Step 1 — choose the positions of the six non-zero entries:
There are $$9$$ positions in a $$3 \times 3$$ matrix. Selecting any $$6$$ of them gives $$\binom{9}{6}=84$$ possibilities.

Step 2 — assign a sign (1 or -1) to each chosen position:
Each of the $$6$$ selected positions can independently be $$1$$ or $$-1$$, giving $$2^{6}=64$$ sign patterns.

Step 3 — multiply the counts:
Total number of admissible matrices $$ = 84 \times 64 = 5376$$.

Hence, the number of matrices $$A \in S$$ satisfying the given condition is $$\mathbf{5376}$$.