JEE Main 2022 · 28 June · morning · Single correct · +4 / −1
If the system of linear equations
2x + 3y - z = - 2
x + y + z = 4
x - y + |λ |z = 4λ - 4
where, λ ∈ R, has no solution, then
- (a) λ = 7
- (b) λ = -7
- (c) λ = 8
- (d) λ² = 1
Answer and solution
Answer (b) λ = -7
Solution
Maths sir · 5 steps
Make the determinant zero first, then check which value of λ makes the right-hand side disagree.
- Coefficient determinant. A system with no solution must have Δ = 0. Write μ = |λ| and expand along the first row.
Δ = |2, 3, -1; 1, 1, 1; 1, -1, μ| = 2(μ + 1) - 3(μ - 1) - 1(-1 - 1) = 7 - μ
- Candidates. Δ = 0 gives |λ| = 7, so λ = 7 or λ = -7. For every other λ the system has a unique solution.
- Find the dependence. With |λ| = 7, the left side of equation 3 is -2 × (eq. 1) + 5 × (eq. 2), because -2(2, 3, -1) + 5(1, 1, 1) = (1, -1, 7). So its right side must be -2(-2) + 5(4) = 24 for the equations to agree.
- Test each value. For λ = 7 the right side is 4(7) - 4 = 24, which matches, so there are infinitely many solutions. For λ = -7 it is 4(-7) - 4 = -32 ≠ 24, so the equations contradict each other.
- So. The system has no solution only for λ = -7, which is option (b).
The idea to keep. Δ = 0 only tells you the system is not uniquely solvable. You still have to test the constants to tell no solution apart from infinitely many solutions.
Answer from the NTA final answer key. Solution by PrepWiser.