Under which one of the following condition does the system of equations
`kx + y + z = k - 1`
`x + ky + z = k - 1`
`x + y + kz = k - 1`
have no solution ?

Options

A. k = 1
B. k `ne` -2
C. k = 1 or k = -2
D. k = -2

Explanation

To determine the condition under which the given system of equations has no solution, we will follow these steps:

1. **Write the system of equations in matrix form**:
The given equations are:
\[ kx + y + z = k - 1 \quad (1) \]
\[ x + ky + z = k - 1 \quad (2) \]
\[ x + y + kz = k - 1 \quad (3) \]
We can express this in the form \( AX = B \), where:
\[ A = \begin{bmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} k - 1 \\ k - 1 \\ k - 1 \end{bmatrix} \]

2. **Find the determinant of matrix A**:
To find the condition for the system to have no solution, we need to calculate the determinant of matrix \( A \) and set it equal to zero:
\[ \text{det}(A) = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix} \]

We can calculate this determinant using the formula for a 3x3 matrix:
\[ \text{det}(A) = k \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} \]

Now calculating the 2x2 determinants:
\[ \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} = k^2 - 1 \]
\[ \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} = k - 1 \]
\[ \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} = 1 - k \]

Substituting these back into the determinant calculation:
\[ \text{det}(A) = k(k^2 - 1) - (k - 1) + (1 - k) \]
Simplifying this gives:
\[ \text{det}(A) = k^3 - k - k + 1 + 1 - k = k^3 - 3k + 2 \]

3. **Set the determinant to zero**:
For the system to have no solution, we set the determinant to zero:
\[ k^3 - 3k + 2 = 0 \]

4. **Factor the polynomial**:
We can factor this polynomial. By testing possible rational roots, we find:
\[ k^3 - 3k + 2 = (k - 1)(k^2 + k - 2) \]
Factoring \( k^2 + k - 2 \) gives:
\[ k^2 + k - 2 = (k - 1)(k + 2) \]
Therefore, we have:
\[ k^3 - 3k + 2 = (k - 1)^2(k + 2) = 0 \]

5. **Find the roots**:
The roots of this equation are:
\[ k - 1 = 0 \quad \Rightarrow \quad k = 1 \]
\[ k + 2 = 0 \quad \Rightarrow \quad k = -2 \]

Thus, the system of equations has no solution when \( k = 1 \) or \( k = -2 \).

### Final Answer:
The system of equations has no solution when \( k = 1 \) or \( k = -2 \).