If the equations `x = ay +z, y = az+x and z = ax + y,(ane0)` are consistent having a non-trivial solution, then `a^2 +3= ....`

Answer

To solve the problem, we need to analyze the system of equations given by:

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

We want to find the condition under which these equations have a non-trivial solution, which means that the determinant of the coefficients must be zero.

### Step 1: Rewrite the equations in standard form

We can rewrite the equations in the form \( Ax + By + Cz = 0 \):

1. Rearranging the first equation gives:
\[ x - ay - z = 0 \quad \Rightarrow \quad 1x - ay - 1z = 0 \]

2. Rearranging the second equation gives:
\[ y - az - x = 0 \quad \Rightarrow \quad -1x + 1y - az = 0 \]

3. Rearranging the third equation gives:
\[ z - ax - y = 0 \quad \Rightarrow \quad -ax - 1y + 1z = 0 \]

### Step 2: Form the coefficient matrix

The system of equations can be represented in matrix form as follows:

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

### Step 3: Calculate the determinant

Let \( D \) be the determinant of the coefficient matrix:

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

We will calculate this determinant using cofactor expansion along the first row:

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

### Step 4: Calculate the 2x2 determinants

1. For the first determinant:
\[ \begin{vmatrix} 1 & -a \\ -1 & 1 \end{vmatrix} = (1)(1) - (-1)(-a) = 1 - a \]

2. For the second determinant:
\[ \begin{vmatrix} -1 & -a \\ -a & 1 \end{vmatrix} = (-1)(1) - (-a)(-a) = -1 - a^2 \]

3. For the third determinant:
\[ \begin{vmatrix} -1 & 1 \\ -a & -1 \end{vmatrix} = (-1)(-1) - (1)(-a) = 1 + a \]

### Step 5: Substitute back into the determinant

Now substituting these values back into the determinant \( D \):

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

Expanding this gives:

\[ D = 1 - a - a - a^3 - 1 - a = -a^3 - 3a \]

### Step 6: Set the determinant to zero for non-trivial solutions

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

\[ -a^3 - 3a = 0 \]

Factoring out \(-a\):

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

### Step 7: Solve for \( a \)

This gives us two cases:

1. \( a = 0 \) (not valid since \( a \neq 0 \))
2. \( a^2 + 3 = 0 \)

Since \( a^2 + 3 \) cannot be zero for real numbers, we conclude that:

\[ a^2 + 3 = 0 \]

### Final Answer

Thus, the condition for the equations to have a non-trivial solution is:

\[ a^2 + 3 = 0 \]