If `a !=b`, then the system of equation `ax + by + bz = 0` `bx + ay + bz = 0` and `bx + by + az = 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 given equations:

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

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

We can express 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: Find the determinant of the coefficient matrix

For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:

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

### Step 3: Calculate the determinant

Using the determinant formula for a \(3 \times 3\) matrix, we have:

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

Calculating the \(2 \times 2\) determinants:

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

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

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

Substituting these back into the determinant \(D\):

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

Simplifying this gives:

\[ D = a^3 - ab^2 - b^2a + b^3 \]

Combining like terms:

\[ D = a^3 + b^3 - 3ab^2 \]

### Step 4: Set the determinant to zero for non-trivial solution

For a non-trivial solution, we set the determinant to zero:

\[ a^3 + b^3 - 3ab^2 = 0 \]

### Step 5: Factor the equation

We can factor \(a^3 + b^3\) as follows:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

Thus, we have:

\[ (a + b)(a^2 - ab + b^2) - 3ab^2 = 0 \]

### Step 6: Analyze the conditions

The equation \(a^2 - ab + b^2 = 0\) has no real solutions unless \(a = b = 0\), which contradicts \(a \neq b\). Therefore, we focus on the condition:

\[ a + b = 0 \implies a = -b \]

### Conclusion

The system of equations will have a non-trivial solution if:

\[ a + 2b = 0 \]