Step 1: Write the determinant of the coefficient matrix.
The coefficient matrix is: \[ \begin{bmatrix} k & 1 & 1 1 & k & 1 1 & 1 & k \end{bmatrix}. \]
The determinant \( D \) of the matrix is: \[ D = \begin{vmatrix} k & 1 & 1 1 & k & 1 1 & 1 & k \end{vmatrix}. \]
Expanding along the first row, we get: \[ D = k \begin{vmatrix} k & 1 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k 1 & 1 \end{vmatrix}. \]
Simplifying the individual 2x2 determinants: \[ D = k(k^2 - 1) - (k - 1) + (1 - k). \]
Now, combining like terms: \[ D = k^3 - k - k + 1 + 1 - k = k^3 - 3k + 2. \]
Step 2: Solve for \( k \) such that \( D = 0 \).
Factorizing the cubic equation: \[ k^3 - 3k + 2 = 0 \quad \implies \quad (k - 1)(k^2 + k - 2) = 0. \]
Next, factorizing further: \[ (k - 1)(k - 1)(k + 2) = 0. \]
Step 3: Identify the condition for no solution.
For the system to have no solution, the determinant \( D \) must equal zero, but the augmented matrix should not have the same rank. This implies that \( k \neq 1 \). Thus, the possible value for \( k \) is \( |k| = 2 \).
Final Answer: \[ \boxed{2} \]
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