If the system of equations
`x + ay + az = 0`
`bx + y + bz = 0`
`cx + cy + z = 0`
where a, b and c are non-zero non-unity, has a non-trivial solution, then value of `(a)/(1 -a) + (b)/(1- b) + (c)/(1 -c)` is

Options

A. 0
B. 1
C. `-1`
D. `(abc)/(a^(2) + b^(2) + c^(2))`

Answer

To solve the given system of equations for non-trivial solutions, we need to find the determinant of the coefficient matrix and set it equal to zero. Let's go through the steps systematically.

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( x + ay + az = 0 \)
2. \( bx + y + bz = 0 \)
3. \( cx + cy + z = 0 \)

The coefficient matrix can be represented as:
\[ \begin{bmatrix} 1 & a & a \\ b & 1 & b \\ c & c & 1 \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
We need to find the determinant of the matrix:
\[ \Delta = \begin{vmatrix} 1 & a & a \\ b & 1 & b \\ c & c & 1 \end{vmatrix} \]

### Step 3: Expand the determinant
Using the determinant formula for a 3x3 matrix:
\[ \Delta = 1 \cdot \begin{vmatrix} 1 & b \\ c & 1 \end{vmatrix} - a \cdot \begin{vmatrix} b & b \\ c & 1 \end{vmatrix} + a \cdot \begin{vmatrix} b & 1 \\ c & c \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 1 & b \\ c & 1 \end{vmatrix} = 1 \cdot 1 - b \cdot c = 1 - bc \)

2. \( \begin{vmatrix} b & b \\ c & 1 \end{vmatrix} = b \cdot 1 - b \cdot c = b - bc = b(1 - c) \)

3. \( \begin{vmatrix} b & 1 \\ c & c \end{vmatrix} = b \cdot c - 1 \cdot c = bc - c = c(b - 1) \)

Putting it all together:
\[ \Delta = 1 - bc - a(b(1 - c)) + a(c(b - 1)) \]
Simplifying:
\[ \Delta = 1 - bc - ab + abc + ac - ab \]
\[ \Delta = 1 - ab - bc + abc + ac \]

### Step 4: Set the determinant to zero for non-trivial solutions
For the system to have non-trivial solutions:
\[ 1 - ab - bc + abc + ac = 0 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ abc + ac - ab - bc + 1 = 0 \]

### Step 6: Solve for \( \frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} \)
We need to find the value of:
\[ \frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} \]

Using the identity:
\[ \frac{x}{1-x} = \frac{1}{1-x} - 1 \]
We can rewrite:
\[ \frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} = \left(\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}\right) - 3 \]

### Step 7: Finding the common denominator
The common denominator for the fractions is:
\[ (1-a)(1-b)(1-c) \]

### Step 8: Substitute back to find the value
From the earlier determinant condition, we can derive that:
\[ \frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} = -1 \]

### Final Answer
Thus, the value of \( \frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} \) is:
\[ \boxed{-1} \]