Question

Let the following system of equations: $$ kx + y + z = 1, \quad x + ky + z = k, \quad x + y + kz = k^2 $$ have no solution. Find $|k|$:

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( 3 \)

Hint:

When determining parameter values in a system of equations, apply properties of determinants and verify consistency using the rank of the augmented matrix.

Answer 1

The Correct Option is C

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: \[ \boxed{2} \]