JEE Main2018MathematicsDeterminantsActual
Let S be the set of all real values of k for which the system of linear equations x + y + z = 2 2 x + y - z = 3 3 x + 2 y + k z = 4 has a unique solution. Then, S is :
Options
- Aequal to R – 0
- Ban empty set
- Cequal to R
- Dequal to 0
Correct answer
A. equal to R – 0
Step-by-step solution
For unique solution, ∆ ≠ 0 ⇒ 1 1 1 2 1 - 1 3 2 k ≠ 0 ⇒ 1 k + 2 - 1 2 k + 3 + 1 4 - 3 ≠ 0   ⇒ - k ≠ 0 ⇒ k ≠ 0 Thus, S = R - 0