JEE Main 2023 · 24 January · evening · Single correct · +4 / −1

If the system of equations
x+2y+3z=3
4x+3y-4z=4
8x+4y-λ z=9+μ
has infinitely many solutions, then the ordered pair (λ,μ) is equal to :

  1. (a) ( 72/5,21/5 )
  2. (b) ( - 72/5, - 21/5 )
  3. (c) ( - 72/5,21/5 )
  4. (d) ( 72/5, - 21/5 )
Answer and solution

Answer (d) ( 72/5, - 21/5 )

Solution

Maths sir · 5 steps

Infinitely many solutions here means the third equation must be built from the first two. Find that mix.

  1. What infinite means. The first two equations are independent, since their x and y coefficients are not in proportion. For infinitely many solutions, the third equation must be a combination a(eq 1) + b(eq 2) of the first two, with the constant term matching too.

    a(x+2y+3z-3) + b(4x+3y-4z-4) = 8x+4y-λ z-(9+μ)

  2. Match x and y. From x: a + 4b = 8. From y: 2a + 3b = 4. Put a = 8 - 4b into the second to get 16 - 5b = 4, so b = 12/5 and a = 8 - 48/5 = -8/5.
  3. Match z. 3a - 4b = -24/5 - 48/5 = -72/5. This must equal -λ, so λ = 72/5. As a check, the coefficient determinant comes out to 5λ - 72, which is 0 here.
  4. Match constants. 3a + 4b = -24/5 + 48/5 = 24/5. This must equal 9 + μ, so μ = 24/5 - 45/5 = -21/5.
  5. So. (λ, μ) = (72/5, -21/5), which is option (d).

The idea to keep. For infinitely many solutions in three variables, Δ = 0 alone is not enough. Write the third equation as a mix of the other two, then match the constants as well. That one check fixes the second parameter.

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

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