Answer

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**Hint:**

We first try to find the condition for symmetric matrices. Then we divide the whole matrix into positional ways to assign numbers. We also have the conditions on the entries where five of these entries are 1 and four of them are 0. We find the possible outcomes of the elements and find the number of matrices.

**Complete step by step answer:**

A be the set of all $3\times 3$ symmetric matrices all of whose entries are either 0 or 1.

For symmetric matrices, the condition is ${{a}_{ij}}={{a}_{ji}}$ where ${{a}_{ij}}$ is the element of ${{i}^{th}}$ row and ${{j}^{th}}$ column.

We divide the whole matrix into three parts. Diagonal elements, upper-triangular elements, lower-triangular elements.

We only need to assign elements for Diagonal elements and anyone triangular part as the other one will automatically be placed. So, we only need to care about six elements.

We also have that five of those nine entries are 1 and four of them are 0.

Now we start with possible outcomes. We assign numbers for diagonal elements.

Outcome-1: Diagonal elements are of two 0s and a single 1.

For this setup, we have one 0 and two 1s for the upper-triangular elements. The number of ways these arrangements can be done is ${}^{3}{{C}_{2}}\times {}^{3}{{C}_{2}}=3\times 3=9 $.

Outcome-2: Diagonal elements are all 1s.

For this setup, we have one 1 and two 0s for the upper-triangular elements. The number of ways these arrangements can be done is ${}^{3}{{C}_{2}}=3 $.

Therefore, the total number of outcomes is $ 9+3=12 $. The correct option is A.

**Note:**

We have to remember that one triangular part is decided by the other for the condition of ${{a}_{ij}}={{a}_{ji}}$. If one element changes in a triangular part then the change also happens on the other part. We also need to remember the theorem ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!} $.

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