Given that the system of equations `x=cy+bz ,y=az+cx , z=bx +ay` has nonzero solutions and and at least one of the a,b,c is a proper fraction.
abc is

Options

A. `gt-1`
B. `gt1`
C. `lt2`
D. `lt3`

Explanation

To solve the problem, we need to analyze the system of equations given:

1. \( x = cy + bz \)
2. \( y = az + cx \)
3. \( z = bx + ay \)

We are tasked with finding the value of \( abc \) given that the system has nonzero solutions and at least one of \( a, b, c \) is a proper fraction.

### Step 1: Write the system in matrix form

We can express the system of equations in matrix form as follows:

\[ \begin{pmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]

### Step 2: Find the determinant of the coefficient matrix

For the system to have nonzero solutions, the determinant of the coefficient matrix must be zero:

\[ \text{Det} = \begin{vmatrix} 1 & -c & -b \\ -a & 1 & -a \\ -b & -a & 1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant

We can expand the determinant:

\[ \text{Det} = 1 \cdot \begin{vmatrix} 1 & -a \\ -a & 1 \end{vmatrix} - (-c) \cdot \begin{vmatrix} -a & -a \\ -b & 1 \end{vmatrix} - (-b) \cdot \begin{vmatrix} -a & 1 \\ -b & -a \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 1 & -a \\ -a & 1 \end{vmatrix} = 1 + a^2 \)

2. \( \begin{vmatrix} -a & -a \\ -b & 1 \end{vmatrix} = -a + ab \)

3. \( \begin{vmatrix} -a & 1 \\ -b & -a \end{vmatrix} = a^2 - b \)

Putting it all together:

\[ \text{Det} = (1 + a^2) + c(-a + ab) + b(a^2 - b) = 0 \]

### Step 4: Simplify the determinant equation

Expanding and simplifying gives us:

\[ 1 + a^2 - ac + abc + ab^2 - b^2 = 0 \]

Rearranging leads to:

\[ a^2 + b^2 + c^2 + 2abc = 1 \]

### Step 5: Analyze the inequality

Since we are given that at least one of \( a, b, c \) is a proper fraction, we can derive:

1. \( 1 - a^2 > 0 \)
2. \( 1 - b^2 > 0 \)
3. \( 1 - c^2 > 0 \)

Adding these inequalities gives:

\[ 3 > a^2 + b^2 + c^2 \]

### Step 6: Substitute back into the equation

From \( a^2 + b^2 + c^2 + 2abc = 1 \):

Substituting \( a^2 + b^2 + c^2 < 3 \):

\[ 1 + 2abc < 3 \]

This implies:

\[ 2abc < 2 \implies abc < 1 \]

### Conclusion

Thus, we have established that:

\[ abc > -1 \quad \text{and} \quad abc < 1 \]

So the final answer is:

\[ abc \text{ is a proper fraction.} \]