Total Number of Symmetric Matrices with Specific Binary Entries

To find the number of matrices in the set $\mathcal{A}$, let a general $3 \times 3$ symmetric matrix $A$ be represented as:

$$A = \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \end{pmatrix}$$

where all entries $a, b, c, d, e, f \in \{0, 1\}$.

The matrix consists of:

• $3$ diagonal entries: $a, d, f$, which each appear once in the matrix.
• $3$ pairs of off-diagonal entries: $(b, b)$, $(c, c)$, $(e, e)$, where each pair contributes two identical entries to the matrix.

The total number of $1$'s in the matrix is $5$, so the sum of all elements in the matrix must be $5$:

$$(a + d + f) + 2(b + c + e) = 5$$

Let $S_{\text{diag}} = a + d + f$ and $S_{\text{off}} = b + c + e$.
Since $a, d, f, b, c, e \in \{0, 1\}$, we have:

$$S_{\text{diag}} \in \{0, 1, 2, 3\} \quad \text{and} \quad S_{\text{off}} \in \{0, 1, 2, 3\}$$

The equation becomes:

$$S_{\text{diag}} + 2 S_{\text{off}} = 5$$

Since $2S_{\text{off}}$ is an even integer, $S_{\text{diag}}$ must be an odd integer. Hence, $S_{\text{diag}}$ can only be $1$ or $3$.

Case 1: $S_{\text{diag}} = 1$

$$1 + 2S_{\text{off}} = 5 \implies S_{\text{off}} = 2$$

• Number of ways to choose the diagonal elements $(a, d, f)$ such that their sum is $1$:

$$\binom{3}{1} = 3$$

• Number of ways to choose the off-diagonal elements $(b, c, e)$ such that their sum is $2$:

$$\binom{3}{2} = 3$$

• Total matrices in this case:

$$3 \times 3 = 9$$

Case 2: $S_{\text{diag}} = 3$

$$3 + 2S_{\text{off}} = 5 \implies S_{\text{off}} = 1$$

• Number of ways to choose the diagonal elements $(a, d, f)$ such that their sum is $3$:

$$\binom{3}{3} = 1$$

• Number of ways to choose the off-diagonal elements $(b, c, e)$ such that their sum is $1$:

$$\binom{3}{1} = 3$$

• Total matrices in this case:

$$1 \times 3 = 3$$

Adding the matrices from both cases:

$$\text{Total number of matrices} = 9 + 3 = 12$$

Therefore, the correct option is (A).