Question

If the system of equations \( kx + y + z = k - 1 \), \( x + ky + z = k - 1 \), and \( x + y + kz = k - 1 \) has infinite solutions, then \( k \) is

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

For infinite solutions, check when determinant = 0 and equations are consistent. Plug back to test values.

Answer 1

The Correct Option is B

The given system is: \[ kx + y + z = k - 1 \quad \text{(1)}\\ x + ky + z = k - 1 \quad \text{(2)}\\ x + y + kz = k - 1 \quad \text{(3)} \] Let’s assume the system has infinite solutions. This occurs when the coefficient matrix is singular (i.e., determinant = 0) and the system is consistent. Coefficient matrix: \[ \begin{pmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \\ \end{pmatrix} \] Find its determinant: \[ \text{Det} = k^3 + 2 - 3k \] Set determinant = 0 for infinite solutions: \[ k^3 - 3k + 2 = 0 \Rightarrow (k - 1)^2(k + 2) = 0 \Rightarrow k = 1 \text{ or } -2 \] Now check which value satisfies consistency with the RHS \( = k - 1 \). Try \( k = 1 \): All equations become: \[ x + y + z = 0 \] So the system reduces to 3 identical equations → Infinite solutions. Hence, \( k = 1 \).