JEE Main 2023 · 10 April · morning · Single correct · +4 / −1
For the system of linear equations
2x - y + 3z = 5
3x + 2y - z = 7
4x + 5y + α z = β,
which of the following is NOT correct?
- (a) The system has infinitely many solutions for α=-6 and β=9
- (b) The system has a unique solution for α ≠ -5 and β=8
- (c) The system is inconsistent for α=-5 and β=8
- (d) The system has infinitely many solutions for α=-5 and β=9
Answer and solution
Answer (a) The system has infinitely many solutions for α=-6 and β=9
Solution
Maths sir · 3 steps
Find the α that makes Δ = 0 first; every other α gives a unique solution.
- Coefficient determinant. Expand along the first row. If Δ ≠ 0, the system has exactly one solution, whatever β is.
Δ = |2, -1, 3; 3, 2, -1; 4, 5, α| = 2(2α+5) + (3α+4) + 3(15-8) = 7α + 35
- Zero determinant. Δ = 0 only at α = -5. Then the third equation's left side equals 2 × (second) - (first): 2(3x+2y-z) - (2x-y+3z) = 4x+5y-5z. Its right side must then be 2 × 7 - 5 = 9.
Option Values What happens Statement (a) α=-6, β=9 Δ = -7 ≠ 0, so a unique solution not correct (b) α ≠ -5, β=8 Δ ≠ 0, unique solution correct (c) α=-5, β=8 8 ≠ 9, the equations contradict correct (d) α=-5, β=9 third equation is redundant, infinitely many correct - So. At α = -6 the determinant is not zero, so the system cannot have infinitely many solutions. The incorrect statement is (a).
The idea to keep. For a 3 × 3 system, Δ ≠ 0 means a unique solution, full stop. Only at the α that makes Δ = 0 does β decide between infinitely many and none.
Answer from the NTA final answer key. Solution by PrepWiser.