JEE Main 2023 · 6 April · morning · Single correct · +4 / −1

If the system of equations
x+y+a z=b
2 x+5 y+2 z=6
x+2 y+3 z=3
has infinitely many solutions, then 2 a+3 b is equal to :

  1. (a) 28
  2. (b) 25
  3. (c) 20
  4. (d) 23
Answer and solution

Answer (d) 23

Solution

Maths sir · 4 steps

Infinitely many solutions means the first equation must be built from the other two, so find that combination.

  1. Set D to zero. A system with infinitely many solutions has coefficient determinant D = 0. Expand it along the first row.

    D = |1, 1, a; 2, 5, 2; 1, 2, 3| = 1(15-4) - 1(6-2) + a(4-5) = 7 - a = 0

  2. Find the combination. So a = 7. Look for p, q with p(2, 5, 2) + q(1, 2, 3) = (1, 1, 7). From 2p + q = 1 and 5p + 2q = 1 you get p = -1, q = 3, and the z-check -2 + 9 = 7 works.
  3. Match the constants. The first equation must be 3 × (third) - (second), so its right side must be 3 × 3 - 6 = 3. That gives b = 3. The second and third equations are not multiples of each other, so one variable stays free and there are infinitely many solutions.
  4. So. a = 7 and b = 3, so 2a + 3b = 14 + 9 = 23, which is option (d).

The idea to keep. For infinitely many solutions, D = 0 alone is not enough. The right-hand sides must follow the same combination as the coefficients, so also check D₁ = D₂ = D₃ = 0.

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.