If the system of equations
`x+y+z=0`
`ax+by+z=0`
`bx+y+z=0`
has a non trivial solution then

Options

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

Answer

To determine the condition under which the given system of equations has a non-trivial solution, we need to analyze the coefficient matrix and its determinant. The system of equations is as follows:

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

### Step 1: Write the Coefficient Matrix
The coefficient matrix \( A \) for the given system of equations is:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ a & b & 1 \\ b & 1 & 1 \end{bmatrix} \]

### Step 2: Set Up the Determinant
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be equal to zero:
\[ \text{det}(A) = 0 \]

### Step 3: Calculate the Determinant
We can calculate the determinant using the formula for a 3x3 matrix:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} b & 1 \\ 1 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} a & 1 \\ b & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} a & b \\ b & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} b & 1 \\ 1 & 1 \end{vmatrix} = b \cdot 1 - 1 \cdot 1 = b - 1 \)
2. \( \begin{vmatrix} a & 1 \\ b & 1 \end{vmatrix} = a \cdot 1 - 1 \cdot b = a - b \)
3. \( \begin{vmatrix} a & b \\ b & 1 \end{vmatrix} = a \cdot 1 - b \cdot b = a - b^2 \)

Substituting these back into the determinant expression:
\[ \text{det}(A) = 1(b - 1) - 1(a - b) + 1(a - b^2) \]
\[ = b - 1 - a + b + a - b^2 \]
\[ = 2b - b^2 - 1 \]

### Step 4: Set the Determinant to Zero
Setting the determinant equal to zero for a non-trivial solution:
\[ 2b - b^2 - 1 = 0 \]

### Step 5: Rearranging the Equation
Rearranging gives us:
\[ b^2 - 2b + 1 = 0 \]

### Step 6: Factor the Quadratic
This can be factored as:
\[ (b - 1)^2 = 0 \]

### Step 7: Solve for \( b \)
Thus, we find:
\[ b - 1 = 0 \implies b = 1 \]

### Conclusion
The condition for the system of equations to have a non-trivial solution is:
\[ b = 1 \]