The values of `k` for which the system of equations `kx+y+z=0,x-ky+z=0` and `x+y+z=0` possesses non-trivial solution. Are

Options

A. `1,2`
B. `-1,1`
C. `1,-2`
D. `-1,-2`

Explanation

To find the values of \( k \) for which the system of equations

1. \( kx + y + z = 0 \)
2. \( x - ky + z = 0 \)
3. \( x + y + z = 0 \)

possesses a non-trivial solution, we can use the condition that the determinant of the coefficients must be zero.

### Step-by-Step Solution:

1. **Write the system of equations in matrix form:**
The given equations can be represented in matrix form as:
\[ \begin{bmatrix} k & 1 & 1 \\ 1 & -k & 1 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

2. **Set up the determinant:**
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:
\[ \text{det} \begin{bmatrix} k & 1 & 1 \\ 1 & -k & 1 \\ 1 & 1 & 1 \end{bmatrix} = 0 \]

3. **Calculate the determinant:**
We calculate the determinant using the formula for a \( 3 \times 3 \) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \).

Here, \( a = k, b = 1, c = 1, d = 1, e = -k, f = 1, g = 1, h = 1, i = 1 \).

Thus, the determinant becomes:
\[ = k((-k)(1) - (1)(1)) - 1(1(1) - (1)(1)) + 1(1(1) - (-k)(1)) \]
Simplifying this:
\[ = k(-k - 1) - 1(1 - 1) + 1(1 + k) \]
\[ = k(-k - 1) + 1 + k \]
\[ = -k^2 - k + 1 + k \]
\[ = -k^2 + 1 \]

4. **Set the determinant to zero:**
Now, we set the determinant equal to zero:
\[ -k^2 + 1 = 0 \]
\[ k^2 = 1 \]

5. **Solve for \( k \):**
Taking the square root of both sides gives:
\[ k = \pm 1 \]

### Conclusion:
The values of \( k \) for which the system of equations possesses a non-trivial solution are:
\[ k = 1 \quad \text{and} \quad k = -1 \]