Let `A=(a_(ij))_(3xx3) and B=(b_(ij))_(3xx3)`, where `b_(ij)=(a_(ij)+a_(ji))/(2) `. Number of such matrices A whose elements are selected from the set `{0, 1, 2, 3}` such that `A=B`. Are

Options

A. `2^(9)`
B. `2^(12)`
C. `2^(6)`
D. `2^(8)`

Answer

To solve the problem, we need to determine the number of 3x3 matrices \( A \) such that \( A = B \), where \( B \) is defined by the relationship \( b_{ij} = \frac{a_{ij} + a_{ji}}{2} \).

### Step-by-Step Solution:

1. **Understanding the Condition \( A = B \)**:
Since \( A = B \), we have:
\[ a_{ij} = b_{ij} = \frac{a_{ij} + a_{ji}}{2} \]
This implies:
\[ 2a_{ij} = a_{ij} + a_{ji} \]
Rearranging gives:
\[ a_{ij} = a_{ji} \]
This means that the matrix \( A \) must be symmetric.

2. **Structure of a 3x3 Symmetric Matrix**:
A 3x3 symmetric matrix can be represented as:
\[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \]
Here, the elements \( a_{12}, a_{13}, a_{23} \) are mirrored across the diagonal.

3. **Identifying the Elements**:
The diagonal elements are \( a_{11}, a_{22}, a_{33} \) and the off-diagonal elements are \( a_{12}, a_{13}, a_{23} \). Thus, we have:
- 3 diagonal elements: \( a_{11}, a_{22}, a_{33} \)
- 3 off-diagonal elements: \( a_{12}, a_{13}, a_{23} \)

4. **Choosing Values from the Set**:
Each element of the matrix \( A \) can be chosen from the set \( \{0, 1, 2, 3\} \). Therefore, for each of the 6 positions in the matrix, we have 4 choices.

5. **Calculating the Total Number of Matrices**:
Since there are 6 independent positions in the symmetric matrix and each can take 4 values, the total number of symmetric matrices \( A \) is:
\[ 4^6 \]

6. **Final Calculation**:
We can simplify \( 4^6 \):
\[ 4^6 = (2^2)^6 = 2^{12} \]

Thus, the number of such matrices \( A \) is \( 2^{12} \).

### Conclusion:
The answer is \( 2^{12} \).