The system of linear equations `kx + y + z = 1, x + ky + z = 1 and x + y + kz = 1` has a unique solution under which one of the following conditions ?

Options

A. `k ne 1 and k ne -2`
B. `k ne 1 and k ne 2`
C. `k ne -1 and k ne -2`
D. `k ne -1 and k ne 2`

Explanation

To determine the conditions under which the given system of linear equations has a unique solution, we can represent the system in matrix form and analyze the determinant of the coefficient matrix.

### Step-by-Step Solution:

1. **Write the system of equations:**
The given equations are:
\[ kx + y + z = 1 \quad (1) \]
\[ x + ky + z = 1 \quad (2) \]
\[ x + y + kz = 1 \quad (3) \]

2. **Formulate the coefficient matrix and the constant matrix:**
The coefficient matrix \( A \) and the constant matrix \( B \) can be written as:
\[ A = \begin{bmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \]

3. **Calculate the determinant of the coefficient matrix \( A \):**
To find the conditions for a unique solution, we need to ensure that the determinant of matrix \( A \) is non-zero. The determinant can be calculated as follows:
\[ \text{det}(A) = k \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} \]

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:
\[ = k^3 - k - k + 1 + 1 - k \]
\[ = k^3 - 3k + 2 \]

4. **Set the determinant not equal to zero:**
For the system to have a unique solution, we require:
\[ k^3 - 3k + 2 \neq 0 \]

5. **Find the roots of the equation:**
To find the values of \( k \) that make the determinant zero, we can factor or use the Rational Root Theorem. Testing possible rational roots, we find:
\[ k = 1 \quad \text{and} \quad k = -2 \]
Thus, the polynomial can be factored as:
\[ (k - 1)(k + 2)(k - 1) = 0 \]

6. **Conclusion:**
Therefore, the conditions for the system to have a unique solution are:
\[ k \neq 1 \quad \text{and} \quad k \neq -2 \]