JEE Main 2023 · 6 April · evening · Single correct · +4 / −1
For the system of equations
x+y+z=6
x+2 y+α z=10
x+3 y+5 z=β, which one of the following is NOT true?
- (a) System has a unique solution for α=3,β≠14.
- (b) System has infinitely many solutions for α=3, β=14.
- (c) System has no solution for α=3, β=24.
- (d) System has a unique solution for α=-3, β=14.
Answer and solution
Answer (a) System has a unique solution for α=3,β≠14.
Solution
Maths sir · 3 steps
Set the determinant to zero: at α = 3 a unique solution is impossible, whatever β is.
- Coefficient determinant. The system has a unique solution exactly when the determinant of the coefficients is non-zero. Expand along the first row.
D = |1, 1, 1; 1, 2, α; 1, 3, 5| = (10 - 3α) - (5 - α) + (3 - 2) = 6 - 2α
- When alpha is 3. D = 0, so there is no unique solution. Subtracting the first equation from the other two gives y + 2z = 4 and 2y + 4z = β - 6. These agree only when β - 6 = 8, that is β = 14 (infinitely many solutions), and any other β gives no solution.
Option Values What really happens Statement is (a) α = 3, β ≠ 14 D = 0 and the equations clash: no solution NOT true (b) α = 3, β = 14 D = 0, equations agree: infinitely many true (c) α = 3, β = 24 24 ≠ 14: no solution true (d) α = -3, β = 14 D = 6 + 6 = 12 ≠ 0: unique true - So. With α = 3 the determinant is zero, so the system can never have a unique solution. The false statement is (a).
The idea to keep. First set D = 0 to find the bad parameter value. At that value there is no unique solution. Then check whether the equations agree (infinitely many solutions) or clash (no solution).
Answer from the NTA final answer key. Solution by PrepWiser.