JEE Main 2025 · 2 April · morning · Single correct · +4 / −1

If the system of linear equations
3 x+y+β z=3 ; 2 x+α y-z=-3 ; x+2 y+z=4
has infinitely many solutions, then the value of 22 β-9 α is :

  1. (a) 31
  2. (b) 37
  3. (c) 43
  4. (d) 49
Answer and solution

Answer (a) 31

Solution

Maths sir · 4 steps

Use Cramer's determinants: D_y holds only β and D_z holds only α, so each gives one value.

  1. The condition. For infinitely many solutions, D = 0 and also Dₓ = D_y = D_z = 0. Pick the determinants where only one unknown survives. Replacing the y-column or the z-column with the constants (3, -3, 4) removes α or β.

    D_y = |3, 3, β; 2, -3, -1; 1, 4, 1| = 3(1) - 3(3) + β(11) = 11β - 6

  2. Find both values. D_y = 0 gives β = 6/11. For D_z, put the constants in the z-column: |3, 1, 3; 2, α, -3; 1, 2, 4| = 3(4α + 6) - 11 + 3(4 - α) = 9α + 19. Setting it to zero gives α = -19/9.
  3. Check D. D = 3α + 3 + 4β - αβ. With these values it is (-209 + 99 + 72 + 38)/33 = 0, and Dₓ = 0 as well. So the system really has infinitely many solutions.
  4. So. 22β - 9α = 22 × 6/11 + 9 × 19/9 = 12 + 19 = 31, which is option (a).

The idea to keep. For infinitely many solutions every Cramer determinant must vanish. Start with the ones that contain a single unknown, and they hand you the parameters one by one.

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 2025.