JEE Main 2020 · 2 September · morning · Single correct · +4 / −1
Let S be the set of all λ ∈ R for which the system of linear equations
2x – y + 2z = 2
x – 2y + λz = –4
x + λy + z = 4
has no solution. Then the set S :
- (a) contains more than two elements.
- (b) contains exactly two elements.
- (c) is a singleton.
- (d) is an empty set.
Answer and solution
Answer (b) contains exactly two elements.
Solution
Maths sir · 4 steps
Find where the determinant vanishes, then check each such λ for a contradiction in the equations.
- When no solution. If the coefficient determinant Δ ≠ 0, the system has exactly one solution. So "no solution" can only happen at values of λ where Δ = 0.
Δ = |2, -1, 2; 1, -2, λ; 1, λ, 1| = -2λ² + λ + 1 = -(2λ + 1)(λ - 1)
- Check λ = 1. Subtracting the second equation from the third gives 3y = 8, so y = 8/3. The first equation then gives x + z = 7/3, but the third gives x + z = 4/3. These contradict each other, so there is no solution.
- Check λ = −1/2. Halving the first equation gives x - y/2 + z = 1, while the third equation says x - y/2 + z = 4. The same expression cannot equal both, so again there is no solution.
- So. S = {1, -1/2}, so S contains exactly two elements. The answer is (b).
The idea to keep. No solution can only occur where Δ = 0. At each root, test whether the equations contradict each other (no solution) or one is a combination of the others (infinitely many).
Answer from the NTA final answer key. Solution by PrepWiser.