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

Options

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

Answer

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

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

has a non-zero solution, we need to set up the corresponding determinant and find when it equals zero.

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

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

### Step 2: Set up the determinant
For the system to have a non-zero solution, the determinant of the coefficient matrix must be 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 the formula for a 3x3 matrix:

\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Substituting the values from our matrix:

\[ \text{Det} = 1 \left((-1)(-1) - (-1)(1)\right) - (-k) \left(k(-1) - (-1)(1)\right) + (-1) \left(k(1) - (-1)(1)\right) \]

Calculating each term:

1. First term: \( 1 \cdot (1 + 1) = 1 \cdot 2 = 2 \)
2. Second term: \( k(k + 1) \)
3. Third term: \( -1(k + 1) = -k - 1 \)

Putting it all together:

\[ \text{Det} = 2 + k(k + 1) - (k + 1) = 0 \]

### Step 4: Simplify the equation
Now we simplify the equation:

\[ 2 + k^2 + k - k - 1 = 0 \]

This simplifies to:

\[ k^2 + 1 = 0 \]

### Step 5: Solve for \( k \)
From \( k^2 + 1 = 0 \):

\[ k^2 = -1 \]

This implies:

\[ k = i \quad \text{or} \quad k = -i \]

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

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