Another dash of symmetry .....

For the matrix $\textbf{A}$ to be symmetric it must be equal to its transpose, i.e., the entries must be such that $a_{ij}=a_{ji}$ .

For a $3$ x $3$ matrix this means that we require the $3$ "pairings" $a_{12}=a_{21}, a_{13}=a_{31}$ and $a_{23}=a_{32}$ , while $a_{11}, a_{22}$ and $a_{33}$ are unchanged by the transposing.

Now we need to look separately at the cases where there are $0, 1, 2, ... ,9$ entries that are $0$ .

There is only $1$ matrix that has no $0$ 's, and this matrix is indeed symmetric.

To have just one $0$ , all the pairing must be $1$ 's and precisely $1$ of the $a_{ii}$ entries must be $0$ . This gives us $3$ symmetric matrices.

We can have $2$ entries that are $0$ in the following ways: (i) $2$ of the $a_{ii}$ entries are $0$ , and all other entries are $1$ , and (ii), precisely one of the pairings is composed of $0$ 's, and all other entries are $1$ 's. Case (i) gives us $\binom{3}{2} = 3$ symmetric matrices, and case (ii) gives us $\binom{3}{1} = 3$ more symmetric matrices, for a total of $6$ .

We can have $3$ entries that are $0$ in the following ways: (i) all the $a_{ii}$ entries are $0$ and all the pairings are $1$ 's, and (ii) one of the pairings is composed of $0$ 's and one of the $a_{ii}$ entries is a $0$ . Case {i} gives us $1$ symmetric matrix, and case (ii) gives us $\binom{3}{1} \binom{3}{1} = 3*3 = 9$ more, for a total of $10$ .

We can have $4$ entries that are $0$ in the following ways: (i) one of the pairings is composed of $0$ 's and $2$ of the $a_{ii}$ entries are $0$ 's, and (ii) $2$ of the pairings are composed of $0$ 's, with all remaining entries being $1$ 's. Case (i) gives us $\binom{3}{1} \binom{3}{2} = 9$ symmetric matrices, and case (ii) gives us $\binom{3}{2} = 3$ more, for a total of $12$ .

The number of symmetric matrices with $5$ entries that are $0$ is the same as the number with $4$ entries that are $1$ , which by symmetry will be the same as the number with $4$ entries that are $0$ , i.e., $12$ . Similarly, the number of symmetric matrices with $6, 7, 8$ and $9$ entries that are $0$ will be $10, 6, 3$ and $1$ , respectively.

The desired sum is then $2*(1^{2} + 3^{2} + 6^{2} + 10^{2} + 12^{2}) = \boxed{580}$ .