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)]`
has a unique solution is

Options

A. less than 4
B. at least 4 but less than 7
C. at least 7 but less than 10
D. at leat 10

Explanation

To solve the problem, we need to find the number of \(3 \times 3\) symmetric matrices with entries either 0 or 1, where five entries are 1 and four are 0. We also need to ensure that the system of equations \(A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) has a unique solution.

### Step 1: Understanding the structure of the symmetric matrix

A \(3 \times 3\) symmetric matrix has the following structure:

\[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \]

This matrix has 6 unique entries because the entries on the diagonal can be chosen independently, while the off-diagonal entries are mirrored.

### Step 2: Counting the total number of entries

In total, the symmetric matrix has 6 entries. We need to choose 5 entries to be 1 and 1 entry to be 0.

### Step 3: Choosing the entries

The total number of ways to choose 5 entries from 6 to be 1 is given by:

\[ \binom{6}{5} = 6 \]

### Step 4: Ensuring a unique solution

For the system \(A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) to have a unique solution, the matrix \(A\) must be invertible. A symmetric matrix is invertible if and only if its determinant is non-zero.

### Step 5: Determining when the determinant is non-zero

To ensure that the determinant is non-zero, we need to avoid configurations that lead to linear dependence among the rows of the matrix.

1. If all three diagonal entries \(a_{11}, a_{22}, a_{33}\) are 1, then the matrix is more likely to be invertible.
2. If any row or column is entirely zeros, the determinant will be zero.

### Step 6: Valid configurations

We can analyze the valid configurations based on the placement of the zeros:

- If we place the 0 in any of the diagonal positions (which are \(a_{11}, a_{22}, a_{33}\)), we will have two diagonal entries as 1 and one as 0, which may lead to a non-invertible matrix.
- If the 0 is placed in one of the off-diagonal positions, we can maintain two 1s on the diagonal, which is more likely to yield an invertible matrix.

### Step 7: Counting valid matrices

1. Place the 0 in one of the off-diagonal positions (3 choices: \(a_{12}, a_{13}, a_{23}\)).
2. The remaining 5 entries can be filled with 1s, ensuring that the diagonal entries remain 1.

Thus, the total number of valid matrices is 3.

### Conclusion

The number of \(3 \times 3\) symmetric matrices \(A\) for which the system of linear equations \(A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) has a unique solution is:

\[ \boxed{3} \]