The system of equations ax + y + z = 0 , -x + ay + z = 0 and - x - y + az = 0 has a non-zero solution if the real value of 'a' is

Options

A. 1
B. `-1`
C. 3
D. 0

Answer

To determine the real value of 'a' for which the system of equations has a non-zero solution, we will analyze the given equations and find the determinant of the coefficient matrix. The system of equations is:

1. \( ax + y + z = 0 \)
2. \( -x + ay + z = 0 \)
3. \( -x - y + az = 0 \)

We can represent this system in matrix form as:

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

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

### Step 1: Calculate the Determinant

The determinant of the matrix is given by:

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

### Step 2: Expand the Determinant

Using the first row to expand the determinant, we have:

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

Calculating the 2x2 determinants:

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

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

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

Substituting these back into the determinant:

\[ D = a(a^2 + 1) - 1(-a + 1) + 1(1 - a) \]
\[ = a^3 + a - a + 1 + 1 - a \]
\[ = a^3 + 3a + 1 \]

### Step 3: Set the Determinant to Zero

For the system to have a non-zero solution, we set the determinant equal to zero:

\[ a^3 + 3a + 1 = 0 \]

### Step 4: Solve the Equation

To find the real values of 'a', we can analyze the cubic equation \( a^3 + 3a + 1 = 0 \).

Using the Rational Root Theorem or numerical methods, we can find that \( a = -1 \) is a root.

### Step 5: Factor the Polynomial

We can factor \( a^3 + 3a + 1 \) as:

\[ (a + 1)(a^2 - a + 1) = 0 \]

### Step 6: Analyze the Quadratic Factor

The quadratic \( a^2 - a + 1 \) has no real roots (its discriminant \( (-1)^2 - 4 \cdot 1 \cdot 1 = -3 < 0 \)). Thus, the only real solution is:

\[ a + 1 = 0 \implies a = -1 \]

### Conclusion

The real value of 'a' for which the system of equations has a non-zero solution is:

\[ \boxed{-1} \]