JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
The values of λ and μ such that the system of equations x + y + z = 6, 3x + 5y + 5z = 26, x + 2y + λ z = μ has no solution, are :
- (a) λ = 3, μ = 5
- (b) λ = 3, μ ≠ 10
- (c) λ ≠ 2, μ = 10
- (d) λ = 2, μ ≠ 10
Answer and solution
Answer (d) λ = 2, μ ≠ 10
Solution
Maths sir · 5 steps
First make the determinant zero to find λ, then check which μ makes the equations contradict each other.
- When no solution. If the coefficient determinant D ≠ 0, the system always has exactly one solution. So for no solution we need D = 0 and equations that contradict each other.
D = |1, 1, 1; 3, 5, 5; 1, 2, λ| = (5λ - 10) - (3λ - 5) + (6 - 5) = 2λ - 4
- Find lambda. D = 0 gives 2λ - 4 = 0, so λ = 2.
- Check the equations. Subtract 3 times the first equation from the second: 2y + 2z = 8, so y + z = 4. With λ = 2, subtract the first equation from the third: y + z = μ - 6.
- Spot the clash. Both say what y + z equals. If μ - 6 = 4, that is μ = 10, they agree and there are infinitely many solutions. If μ ≠ 10, they contradict each other.
- So. The system has no solution when λ = 2 and μ ≠ 10, which is option (d).
The idea to keep. For no solution, D = 0 fixes the parameter in the coefficients. Then reduce the equations: equal right sides give infinitely many solutions, different right sides give none.
Answer from the NTA final answer key. Solution by PrepWiser.