Let $$A$$ be a $$3 \times 3$$ matrix having entries from the set $$\{-1, 0, 1\}$$. The number of all such matrices $$A$$ having sum of all the entries equal to $$5$$, is ______.

We need to count $$3 \times 3$$ matrices with entries from $$\{-1, 0, 1\}$$ whose entries sum to $$5$$.

Let $$k$$ be the number of entries equal to 1, $$m$$ be the number equal to $$-1$$, and $$9 - k - m$$ the number equal to 0. We require $$k - m = 5$$ and $$k + m \le 9$$ with $$k, m \ge 0$$. Since $$k = m + 5$$, substituting yields $$2m + 5 \le 9$$, hence $$m \le 2$$.

When $$m = 0$$, it follows that $$k = 5$$ and there are 4 zeros. Choosing 5 positions for 1’s among 9 gives $$\binom{9}{5} = 126$$.

For $$m = 1$$, we have $$k = 6$$ and 2 zeros; thus we choose 6 positions for 1’s and 1 of the remaining 3 positions for a $$-1$$, giving $$\binom{9}{6} \times \binom{3}{1} = 84 \times 3 = 252$$.

When $$m = 2$$, then $$k = 7$$ and there are no zeros, so we select 7 positions for 1’s (the remaining 2 are $$-1$$’s), yielding $$\binom{9}{7} = 36$$.

Adding these counts gives $$126 + 252 + 36 = 414$$, and therefore the total number of such matrices is $$414$$.