JEE Main 2021 · 1 September · evening · Single correct · +4 / −1
Consider the system of linear equations
-x + y + 2z = 0
3x - ay + 5z = 1
2x - 2y - az = 7
Let S₁ be the set of all a∈R for which the system is inconsistent and S₂ be the set of all a∈R for which the system has infinitely many solutions. If n(S₁) and n(S₂) denote the number of elements in S₁ and S₂ respectively, then
- (a) n(S₁) = 2, n(S₂) = 2
- (b) n(S₁) = 1, n(S₂) = 0
- (c) n(S₁) = 2, n(S₂) = 0
- (d) n(S₁) = 0, n(S₂) = 2
Answer and solution
Answer (c) n(S₁) = 2, n(S₂) = 0
Solution
Maths sir · 4 steps
Find the values of a that make the determinant zero, then test each one back in the equations.
- Coefficient determinant. If Δ ≠ 0, the system has exactly one solution, so that value of a is in neither set. Only the values that make Δ = 0 need checking.
Δ = |-1, 1, 2; 3, -a, 5; 2, -2, -a| = -(a²+10) + (3a+10) + 2(2a-6) = -(a-3)(a-4)
- Check a = 3. Put u = x - y. The first equation gives u = 2z. The second becomes 3u + 5z = 11z = 1, so z = 1/11. The third becomes 2u - 3z = z = 7, which contradicts z = 1/11. No solution.
- Check a = 4. The left side of the third equation, 2x - 2y - 4z, is -2 times the left side of the first, so it must equal -2 × 0 = 0. But the third equation says it equals 7. No solution.
- So. Both a = 3 and a = 4 make the system inconsistent, and no value gives infinitely many solutions. So S₁ = {3, 4} and S₂ is empty, which gives n(S₁) = 2 and n(S₂) = 0, option (c).
The idea to keep. A zero determinant only tells you the solution is not unique. Always put each such value back into the equations to decide between no solution and infinitely many.
Answer from the NTA final answer key. Solution by PrepWiser.