The number of all $$3 \times 3$$ matrices A, with entries from the set $$\{-1, 0, 1\}$$ such that the sum of the diagonal elements of $$AA^T$$ is 3, is

We begin by recalling a standard fact from linear algebra: for any real matrix $$A$$, the trace of the matrix product $$AA^T$$ is equal to the sum of the squares of all the entries of $$A$$. Symbolically, we write

$$\operatorname{tr}(AA^T)=\sum_{i=1}^{3}\sum_{j=1}^{3}a_{ij}^2,$$

where $$A=(a_{ij})$$ is our $$3\times 3$$ matrix. This happens because the diagonal entry in the $$i$$-th row and $$i$$-th column of $$AA^T$$ is $$\sum_{j=1}^{3}a_{ij}^2$$, and adding the three diagonal entries together simply gathers every $$a_{ij}^2$$ once.

Now the question demands that the sum of the diagonal elements of $$AA^T$$ be $$3$$. Using the equality just stated, we translate that requirement into a very concrete arithmetic condition:

$$\sum_{i=1}^{3}\sum_{j=1}^{3}a_{ij}^2=3.$$

Next, we look at the allowed values of each entry $$a_{ij}$$. Every entry must come from the set $$\{-1,0,1\}$$. We immediately notice

$$(-1)^2 = 1,\qquad 0^2 = 0,\qquad 1^2 = 1.$$

So each individual square is either $$0$$ or $$1$$, and never anything else. Therefore, the total sum of all nine squares can only equal the count of how many entries are non-zero. To make that sum equal to $$3$$, we must have

• exactly three entries of $$A$$ equal to either $$1$$ or $$-1$$ (since each such entry contributes $$1$$ to the sum), and
• the remaining six entries equal to $$0$$ (since $$0^2=0$$ contributes nothing).

Thus, our task converts to a purely combinatorial counting problem: choose any three positions in the $$3\times3$$ grid to hold the non-zero numbers, and then decide independently for each chosen position whether the non-zero number is $$1$$ or $$-1$$.

First, we count the ways to choose the positions. There are $$9$$ positions altogether, and we want any $$3$$ of them, so by the binomial coefficient formula we have

$$\binom{9}{3}= \frac{9\times8\times7}{3\times2\times1}=84.$$

Second, after fixing those three positions, each one can carry either $$1$$ or $$-1$$. The two choices are independent for the three spots, giving in total

$$2^3 = 8$$

sign assignments.

Finally, by the multiplication principle of counting, we multiply the two independent counts:

$$\binom{9}{3}\times 2^3 = 84 \times 8 = 672.$$

Therefore, there are exactly $$672$$ matrices satisfying the given condition.

So, the answer is $$672$$.