JEE Main 2021 · 27 August · evening · Single correct · +4 / −1
Let [λ] be the greatest integer less than or equal to λ. The set of all values of λ for which the system of linear equations
x + y + z = 4,
3x + 2y + 5z = 3,
9x + 4y + (28 + [λ])z = [λ] has a solution is :
- (a) R
- (b) (-∞, -9) ∪ (-9, ∞)
- (c) [-9, -8)
- (d) (-∞, -9) ∪ [-8, ∞)
Answer and solution
Answer (a) R
Solution
Maths sir · 4 steps
When the determinant is zero, do not stop there: check whether the equations still agree.
- Name the integer. Let n = [λ], which is an integer. Expanding the coefficient determinant along the first row gives (36 + 2n) - (39 + 3n) + (12 - 18).
D = |1, 1, 1; 3, 2, 5; 9, 4, 28+n| = -(n+9)
- D not zero. If n ≠ -9, then D ≠ 0 and the system has exactly one solution. That covers every λ outside [-9, -8).
- D equals zero. If n = -9, the third equation is 9x + 4y + 19z = -9. Multiply the first equation by -6 and the second by 5, then add: you get 9x + 4y + 19z = -24 + 15 = -9. So the third equation repeats the first two, and the system has infinitely many solutions.
- So. The system has a solution for every real λ: one solution when [λ] ≠ -9 and infinitely many when [λ] = -9. The answer is (a) ℝ.
The idea to keep. D ≠ 0 means exactly one solution. D = 0 only tells you there is no single solution: check whether one equation is a combination of the others, LHS and RHS together.
Answer from the NTA final answer key. Solution by PrepWiser.