If x,y,z are not all zeros and `ax+ y +z=0, x+by +z=0, x + y + cz=0` then `1/(1-a)+1/(1-b)+(1)/(1-c)=`

Options

A. 1
B. `-1`
C. 2
D. 1

Answer

To solve the problem, we need to find the value of \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \) given the equations:

1. \( ax + y + z = 0 \)
2. \( x + by + z = 0 \)
3. \( x + y + cz = 0 \)

where \( x, y, z \) are not all zero.

### Step 1: Set up the determinant

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

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

This determinant must equal zero because the system has non-trivial solutions (not all \( x, y, z \) are zero).

### Step 2: Calculate the determinant

Expanding the determinant using the first row:

\[ D = a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

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

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

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

Substituting these back into the determinant:

\[ D = a(bc - 1) - (c - 1) + (1 - b) \]

Simplifying this, we have:

\[ D = abc - a - c + 1 + 1 - b = abc - a - b - c + 2 \]

Setting this equal to zero gives us:

\[ abc - a - b - c + 2 = 0 \implies abc = a + b + c - 2 \]

### Step 3: Find \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \)

Now, we need to compute:

\[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \]

Finding a common denominator:

\[ \frac{(1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b)}{(1-a)(1-b)(1-c)} \]

### Step 4: Expand the numerator

Expanding the numerator:

1. \( (1-b)(1-c) = 1 - b - c + bc \)
2. \( (1-a)(1-c) = 1 - a - c + ac \)
3. \( (1-a)(1-b) = 1 - a - b + ab \)

Adding these together:

\[ (1 - b - c + bc) + (1 - a - c + ac) + (1 - a - b + ab) \]

Combining like terms:

\[ 3 - 2(a + b + c) + (ab + ac + bc) \]

### Step 5: Substitute \( abc = a + b + c - 2 \)

Substituting \( abc = a + b + c - 2 \) into the denominator:

\[ (1-a)(1-b)(1-c) = 1 - (a + b + c) + (ab + ac + bc) - abc \]

Substituting \( abc \):

\[ = 1 - (a + b + c) + (ab + ac + bc) - (a + b + c - 2) \]

This simplifies to:

\[ = 3 - (a + b + c) + (ab + ac + bc) \]

### Step 6: Final result

Thus, we have:

\[ \frac{3 - 2(a + b + c) + (ab + ac + bc)}{3 - (a + b + c) + (ab + ac + bc)} = 1 \]

So, the final answer is:

\[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \]