If `a ne b`, then the system of equations `ax+by+bz=0, bx + ay + bz = 0, bx + by + ax = 0` will have a non– trivial solution if

Options

A. `a + b = 0`
B. `a + 2b = 0`
C. `2a + b = 0`
D. `a + 4b = 0`

Answer

To determine the conditions under which the system of equations has a non-trivial solution, we need to analyze the following equations:

1. \( ax + by + bz = 0 \)
2. \( bx + ay + bz = 0 \)
3. \( bx + by + ax = 0 \)

### Step 1: Form the Coefficient Matrix
We can represent the system of equations in matrix form as follows:

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

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

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

### Step 3: Calculate the Determinant
We can calculate the determinant \( D \) using the formula for a 3x3 determinant:

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

Calculating the 2x2 determinants:

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

2. \( \begin{vmatrix} b & b \\ b & a \end{vmatrix} = ba - bb = b(a - b) \)

3. \( \begin{vmatrix} b & a \\ b & b \end{vmatrix} = bb - ab = b(b - a) \)

Now substituting these back into \( D \):

\[ D = a(a^2 - b^2) - b[b(a - b)] + b[b(b - a)] \]

### Step 4: Simplify the Determinant
Substituting the values we calculated:

\[ D = a(a^2 - b^2) - b^2(a - b) + b^2(b - a) \]

Notice that the last two terms cancel each other out:

\[ D = a(a^2 - b^2) \]

### Step 5: Set the Determinant Equal to Zero
For a non-trivial solution, we set the determinant to zero:

\[ a(a^2 - b^2) = 0 \]

This gives us two conditions:

1. \( a = 0 \)
2. \( a^2 - b^2 = 0 \) which simplifies to \( a = b \) or \( a = -b \)

### Step 6: Consider the Condition \( a \neq b \)
Since the problem states that \( a \neq b \), we discard \( a = b \). Therefore, we are left with the condition:

\[ a + 2b = 0 \]

### Conclusion
Thus, the system of equations will have a non-trivial solution if:

\[ a + 2b = 0 \]