The system of equations `kx+(k+2)y+(k-2)z=0, (k+2)x+ky+(k+4)z=0 (k-2)x+(k+4)y+kz=0` has a non - trivial solution for

Options

A. exactly one real value of k
B. exactly two real values of k
C. infinite real values of k
D. no real value of k

Answer

To determine the value of \( k \) for which the system of equations has a non-trivial solution, we need to analyze the given homogeneous system of equations:

\[ \begin{align*} 1. & \quad kx + (k + 2)y + (k - 2)z = 0 \\ 2. & \quad (k + 2)x + ky + (k + 4)z = 0 \\ 3. & \quad (k - 2)x + (k + 4)y + kz = 0 \end{align*} \]

### Step 1: Formulate the coefficient matrix
The system can be represented in matrix form as:

\[ \begin{bmatrix} k & k + 2 & k - 2 \\ k + 2 & k & k + 4 \\ k - 2 & k + 4 & k \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

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

\[ \text{det} \begin{bmatrix} k & k + 2 & k - 2 \\ k + 2 & k & k + 4 \\ k - 2 & k + 4 & k \end{bmatrix} = 0 \]

### Step 3: Calculate the determinant
We will calculate the determinant using cofactor expansion.

1. Expand along the first row:

\[ \text{det} = k \cdot \text{det} \begin{bmatrix} k & k + 4 \\ k + 4 & k \end{bmatrix} - (k + 2) \cdot \text{det} \begin{bmatrix} k + 2 & k + 4 \\ k - 2 & k \end{bmatrix} + (k - 2) \cdot \text{det} \begin{bmatrix} k + 2 & k \\ k - 2 & k + 4 \end{bmatrix} \]

2. Calculate the 2x2 determinants:

\[ \text{det} \begin{bmatrix} k & k + 4 \\ k + 4 & k \end{bmatrix} = k^2 - (k + 4)(k + 4) = k^2 - (k^2 + 8k + 16) = -8k - 16 \]

\[ \text{det} \begin{bmatrix} k + 2 & k + 4 \\ k - 2 & k \end{bmatrix} = (k + 2)k - (k + 4)(k - 2) = k^2 + 2k - (k^2 + 2k - 8) = 8 \]

\[ \text{det} \begin{bmatrix} k + 2 & k \\ k - 2 & k + 4 \end{bmatrix} = (k + 2)(k + 4) - k(k - 2) = k^2 + 6k + 8 - (k^2 - 2k) = 8k + 8 \]

### Step 4: Substitute back into the determinant
Substituting these values back into the determinant expression:

\[ \text{det} = k(-8k - 16) - (k + 2)(8) + (k - 2)(8k + 8) \]

Expanding this gives:

\[ \text{det} = -8k^2 - 16k - 8k - 16 + 8k^2 + 8k - 16 = -16k - 16 \]

### Step 5: Set the determinant to zero
Now we set the determinant to zero:

\[ -16k - 16 = 0 \]

### Step 6: Solve for \( k \)
Solving for \( k \):

\[ -16k = 16 \implies k = -1 \]

### Conclusion
Thus, the value of \( k \) for which the system has a non-trivial solution is:

\[ \boxed{-1} \]