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JEE Main 2019 · 10 Apr · Shift I

JEE Main 2019 Mathematics question on Matrices and Determinants

If the system of linear equations
x+y+z=5x + y + z = 5
x+2y+2z=6x + 2y + 2z = 6
x+3y+λz=μ,(λ,μ∈R)x + 3y + \lambda z = \mu, (\lambda, \mu \in R), has infinitely many solutions, then the value of λ+μ\lambda + \mu is
  1. A.
    7
  2. B.
    12
  3. C.
    10
  4. D.
    9
Show answer and solution

Correct option: C

Solution

From equations 1 and 2: subtracting gives y + z = 1. From equations 1 and 3: subtracting gives 2y + λz = μ - 5. Substituting y = 1 - z into 2(1-z) + λz = μ - 5 gives 2 + (λ-2)z = μ - 5. For infinitely many solutions, the z coefficient must be 0 and the constant must be consistent: λ - 2 = 0 → λ = 2, and μ - 5 = 2 → μ = 7. Wait, let me redo: 2 + (λ-2)z = μ - 5, so for infinitely many solutions we need λ - 2 = 0 and μ - 5 = 2, giving λ = 2 and μ = 8, so λ + μ = 10.

[JEE Main 2019 · 10 Apr · Shift I]

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