If the system of equations x-ky-z=0 , kx-y-z=0 , x+y-z=0 has a non-zero solution , then possible values of k are :

Options

A. `-1,2`
B. `1,2`
C. `0,1`
D. `-1,1`

Answer

To solve the problem, we need to determine the values of \( k \) for which the system of equations has a non-zero solution. The equations given are:

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

### Step 1: Write the system in matrix form
We can represent the system of equations in matrix form as follows:

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

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

\[ \text{Det} = \begin{vmatrix} 1 & -k & -1 \\ k & -1 & -1 \\ 1 & 1 & -1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
We can calculate the determinant using cofactor expansion. Let's expand along the first row:

\[ \text{Det} = 1 \cdot \begin{vmatrix} -1 & -1 \\ 1 & -1 \end{vmatrix} - (-k) \cdot \begin{vmatrix} k & -1 \\ 1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} k & -1 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

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

Substituting these back into the determinant:

\[ \text{Det} = 1 \cdot 2 + k(1 - k) - (k + 1) = 2 + k - k^2 - k - 1 = 2 - k^2 - 1 = 1 - k^2 \]

### Step 4: Set the determinant to zero
Now, we set the determinant equal to zero:

\[ 1 - k^2 = 0 \]

### Step 5: Solve for \( k \)
Rearranging gives:

\[ k^2 = 1 \]

Taking the square root of both sides, we find:

\[ k = \pm 1 \]

### Conclusion
The possible values of \( k \) for which the system has a non-zero solution are:

\[ k = 1 \quad \text{and} \quad k = -1 \]