JEE Main 2023 · 25 January · morning · Single correct · +4 / −1

Let S₁ and S₂ be respectively the sets of all a ∈ ℝ - { 0} for which the system of linear equations
ax + 2ay - 3az = 1
(2a + 1)x + (2a + 3)y + (a + 1)z = 2
(3a + 5)x + (a + 5)y + (a + 2)z = 3
has unique solution and infinitely many solutions. Then

  1. (a) n(S₁) = 2 and S₂ is an infinite set
  2. (b) S₁ = Φ and S₂ = ℝ - { 0}
  3. (c) S₁ = ℝ - { 0} and S₂ = Φ
  4. (d) S₁ is an infinite set and n(S₂) = 2
Answer and solution

Answer (c) S₁ = ℝ - { 0} and S₂ = Φ

Solution

Maths sir · 4 steps

Expand the coefficient determinant fully: if it can never be zero, every allowed a gives a unique solution.

  1. Take out a. The system has a unique solution exactly when the coefficient determinant D ≠ 0. Row 1 is a, 2a, -3a, so take a out of it to get D = a Δ, where row 1 of Δ is 1, 2, -3.
  2. Clear row 1. Apply C₂ → C₂ - 2C₁ and C₃ → C₃ + 3C₁. Row 1 becomes 1, 0, 0, row 2 becomes 2a+1, 1-2a, 7a+4 and row 3 becomes 3a+5, -5a-5, 10a+17.

    Δ = (1-2a)(10a+17) + (7a+4)(5a+5) = 15a² + 31a + 37

  3. Check for roots. The discriminant of 15a² + 31a + 37 is 31² - 4 × 15 × 37 = 961 - 2220 = -1259 < 0. The leading coefficient is positive, so Δ > 0 for every real a.
  4. So. D = a(15a² + 31a + 37) ≠ 0 for every a ≠ 0. Every allowed a gives a unique solution, so S₁ = ℝ - {0} and S₂ = Φ, which is option (c).

The idea to keep. For a square system, first find when D = 0. If D never vanishes on the allowed values, the system always has a unique solution, so infinitely many or no solutions are impossible.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Determinants

Asked about 190 times in JEE Main since 2020. This question is from 2023.