Let A be the set of all `3xx3` symmetric matrices all of whose either 0 or 1. Five of these entries are 1 and four of them are 0.
The number of matrices A in A for which the system of linear equations
`A[(x),(y),(z)]=[(1),(0),(0)]`
is inconsistent is

Options

A. 0
B. more than 2
C. 2
D. 1

Explanation

To solve the problem, we need to find the number of symmetric \(3 \times 3\) matrices with entries either 0 or 1, where 5 entries are 1 and 4 entries are 0, such that the system of linear equations \(A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) is inconsistent.

### Step-by-Step Solution:

1. **Understanding the Structure of the Matrix:**
A symmetric \(3 \times 3\) matrix \(A\) has the form:
\[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \]
where \(a_{ij} \in \{0, 1\}\).

2. **Counting the Entries:**
We have 5 entries as 1 and 4 entries as 0. Since the matrix is symmetric, the diagonal entries \(a_{11}, a_{22}, a_{33}\) can each be 1 or 0, and the off-diagonal entries \(a_{12}, a_{13}, a_{23}\) must also be chosen accordingly.

3. **Defining Variables:**
Let \(x\) be the number of 1s on the diagonal and \(y\) be the number of 1s in the off-diagonal positions. The total number of 1s is given by:
\[ x + y = 5 \]
The maximum number of 1s on the diagonal \(x\) can be 3 (since there are 3 diagonal positions). Thus, the possible pairs \((x, y)\) can be:
- \(x = 3, y = 2\)
- \(x = 2, y = 3\)
- \(x = 1, y = 4\)

4. **Case Analysis:**
- **Case 1:** \(x = 3, y = 2\)
- The diagonal entries are all 1, and we need to choose 2 out of the 3 off-diagonal entries to be 1.
- The number of ways to choose 2 from 3 is \(\binom{3}{2} = 3\).

- **Case 2:** \(x = 2, y = 3\)
- We need to choose 2 diagonal entries to be 1 and all 3 off-diagonal entries to be 1.
- The number of ways to choose 2 from 3 is \(\binom{3}{2} = 3\).

- **Case 3:** \(x = 1, y = 4\)
- We need to choose 1 diagonal entry to be 1 and all 3 off-diagonal entries to be 1.
- The number of ways to choose 1 from 3 is \(\binom{3}{1} = 3\).

5. **Total Matrices:**
The total number of symmetric matrices is:
\[ 3 + 3 + 3 = 9 \]

6. **Finding Inconsistent Matrices:**
A system \(A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) is inconsistent if the determinant of the matrix \(A\) is zero. We need to check the determinants for each case.

After checking the determinants for each of the matrices formed in the cases, we find that:
- The matrices from Case 1 and Case 2 yield determinants that are not equal to zero.
- The matrices from Case 3 yield determinants that are equal to zero.

7. **Final Count of Inconsistent Matrices:**
From our analysis, we find that there are 6 matrices that are inconsistent.

### Conclusion:
The number of matrices \(A\) in \(A\) for which the system of linear equations \(A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) is inconsistent is **6**.