JEE Main 2022 · 29 June · morning · Single correct · +4 / −1
If the system of linear equations
2x + y - z = 7
x - 3y + 2z = 1
x + 4y + δz = k, where δ, k ∈ R has infinitely many solutions, then δ + k is equal to:
- (a) -3
- (b) 3
- (c) 6
- (d) 9
Answer and solution
Answer (b) 3
Solution
Maths sir · 5 steps
Infinitely many solutions means the third equation is just a mix of the first two. Find that mix.
- Determinant must vanish. For infinitely many solutions the coefficient determinant must be zero. Expand along the first row.
|2, 1, -1; 1, -3, 2; 1, 4, δ| = 2(-3δ - 8) - 1(δ - 2) - 1(4 + 3) = -7δ - 21 = 0
- Find delta. -7δ - 21 = 0 gives δ = -3.
- Combine the equations. Take a times the first equation plus b times the second to match the left side x + 4y - 3z. The x and y coefficients give 2a + b = 1 and a - 3b = 4, so a = 1, b = -1. Check z: -a + 2b = -3, which matches.
- Find k. The right sides must combine the same way, so k = 1 × 7 + (-1) × 1 = 6. Any other k would make the system inconsistent.
- So. δ + k = -3 + 6 = 3, which is option (b).
The idea to keep. For infinitely many solutions, Δ = 0 fixes the coefficient, and the constant must follow the same combination of equations. Otherwise the system has no solution.
Answer from the NTA final answer key. Solution by PrepWiser.