If a system of equation `– ax + y + z = 0`
`x -by + z =0`
`x + y - cz = 0 ( a,b,c ne -1)` has a non-zero solution then `(1)/(1+ a) + (1)/(1 + b) + (1)/(1 + c) = `

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the given system of equations and find the value of \(\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}\), we will follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \(-ax + y + z = 0\)
2. \(x - by + z = 0\)
3. \(x + y - cz = 0\)

We can represent this system in matrix form as follows:

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

### Step 2: Find the determinant of the coefficient matrix
For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero:

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

Calculating the determinant \(D\):

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

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 \(D\):

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

Simplifying:

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

Setting the determinant to zero for a non-zero solution:

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

### Step 3: Rearranging the equation
Rearranging gives:

\[ abc = a + b - c \]

### Step 4: Find the value of \(\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}\)
Now, we need to compute:

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

Finding a common denominator:

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

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:

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

Thus, the numerator becomes:

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

### Step 5: Final expression
The denominator is:

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

Using the relation \(abc = a + b - c\):

Substituting this into the denominator:

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

This simplifies to:

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

### Conclusion
Thus, we have:

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

Therefore, the final answer is:

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