Step 1: Understand the properties of the matrices in set A.

A is a set of 3x3 matrices with the following properties:

1. Symmetric: For any matrix M in A, $M = M^T$. A general 3x3 symmetric matrix has the form: $M = \begin{pmatrix} a & b & c \\ b & e & f \\ c & f & i \end{pmatrix}$ This matrix is determined by 6 independent entries: the 3 diagonal elements (a, e, i) and the 3 upper-triangular off-diagonal elements (b, c, f).

2. Binary Entries: All entries of the matrix are either 0 or 1.

3. Entry Count: The matrix must have exactly five entries that are 1 and four entries that are 0.

Step 2: Formulate an equation based on the number of '1's.

The total number of '1's in the matrix depends on how many '1's are on the diagonal and how many are in the off-diagonal positions.

• Let $d_1$ be the number of '1's on the diagonal (i.e., how many of a, e, i are 1). $0 \le d_1 \le 3$.
• Let $o_1$ be the number of '1's in the unique off-diagonal positions (i.e., how many of b, c, f are 1). $0 \le o_1 \le 3$.

Each '1' on the diagonal contributes one '1' to the total count. Each '1' in an off-diagonal position b, c, or f corresponds to two '1's in the matrix (e.g., if b=1, then $M_{12}$ and $M_{21}$ are both 1). Therefore, the total number of '1's in the matrix is given by: $\text{Total 1s} = d_1 + 2 \times o_1$

We are given that the total number of '1's is 5, so we have the equation: $d_1 + 2o_1 = 5$

Step 3: Find all possible integer solutions for $(d_1, o_1)$.

We need to find non-negative integer solutions for $(d_1, o_1)$ that satisfy $d_1 + 2o_1 = 5$, subject to the constraints $0 \le d_1 \le 3$ and $0 \le o_1 \le 3$.

• If $o_1 = 0$, then $d_1 = 5$. This is not possible as $d_1 \le 3$.
• If $o_1 = 1$, then $d_1 = 5 - 2(1) = 3$. This is a valid solution ($d_1=3, o_1=1$).
• If $o_1 = 2$, then $d_1 = 5 - 2(2) = 1$. This is a valid solution ($d_1=1, o_1=2$).
• If $o_1 = 3$, then $d_1 = 5 - 2(3) = -1$. This is not possible as $d_1 \ge 0$.

So, there are two possible cases.

Step 4: Calculate the number of matrices for each case.

Case 1: $d_1 = 3$ and $o_1 = 1$

• This means all 3 diagonal elements are 1. The number of ways to choose 3 diagonal positions for the '1's out of 3 available is $\binom{3}{3} = 1$.
• This means 1 of the 3 unique off-diagonal elements is 1 (and the other 2 are 0). The number of ways to choose which off-diagonal position gets the '1' is $\binom{3}{1} = 3$.
• The number of matrices in this case is $\binom{3}{3} \times \binom{3}{1} = 1 \times 3 = 3$.

(Check: Number of 1s = 3 (diagonal) + 21 (off-diagonal) = 5. Number of 0s = 0 (diagonal) + 22 (off-diagonal) = 4. Total entries = 9. This is correct.)

Case 2: $d_1 = 1$ and $o_1 = 2$

• This means 1 of the 3 diagonal elements is 1 (and the other 2 are 0). The number of ways to choose which diagonal position gets the '1' is $\binom{3}{1} = 3$.
• This means 2 of the 3 unique off-diagonal elements are 1 (and the other 1 is 0). The number of ways to choose which two off-diagonal positions get the '1's is $\binom{3}{2} = 3$.
• The number of matrices in this case is $\binom{3}{1} \times \binom{3}{2} = 3 \times 3 = 9$.

(Check: Number of 1s = 1 (diagonal) + 22 (off-diagonal) = 5. Number of 0s = 2 (diagonal) + 21 (off-diagonal) = 4. Total entries = 9. This is correct.)

Step 5: Find the total number of matrices.

The total number of matrices in set A is the sum of the numbers of matrices from all possible cases. $\text{Total matrices} = (\text{Number from Case 1}) + (\text{Number from Case 2})$ $\text{Total matrices} = 3 + 9 = 12$

Thus, there are 12 such matrices in the set A.

Comparing with the options: A: 12 B: 6 C: 9 D: 3

The calculated answer matches option A.