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JEE Main 2019 · 10 April · Shift II

JEE Main 2019 Mathematics question on Matrices and Determinants

Let λ\lambda be a real number for which the system of linear equations
x+y+z=6x + y + z = 6, 4x+λy−λz=λ−24x + \lambda y - \lambda z = \lambda - 2 and 3x+2y−4z=−53x + 2y - 4z = -5
has infinitely many solutions. Then λ\lambda is a root of the quadratic equation
  1. A.
    λ2−3λ−4=0\lambda^2 - 3\lambda - 4 = 0
  2. B.
    λ2+3λ−4=0\lambda^2 + 3\lambda - 4 = 0
  3. C.
    λ2−λ−6=0\lambda^2 - \lambda - 6 = 0
  4. D.
    λ2+λ−6=0\lambda^2 + \lambda - 6 = 0
Show answer and solution

Correct option: C

Solution

For infinitely many solutions, the determinant of the coefficient matrix must be zero.

D = |1, 1, 1; 4, λ, -λ; 3, 2, -4|

Expanding: 1(λ·(-4) - (-λ)·2) - 1(4·(-4) - (-λ)·3) + 1(4·2 - λ·3)
= 1(-4λ + 2λ) - 1(-16 + 3λ) + 1(8 - 3λ)
= -2λ - (-16 + 3λ) + (8 - 3λ)
= -2λ + 16 - 3λ + 8 - 3λ
= 24 - 8λ

Setting D = 0 gives λ = 3.

Now verify λ = 3 actually gives infinitely many solutions (not inconsistency) — it does, as checking the augmented matrix confirms consistency. Since λ = 3 must satisfy the given quadratic, substitute into option 3: 9 - 3 - 6 = 0 ✓. Additionally, option 3 factors as (λ - 3)(λ + 2) = 0, giving roots λ = 3 and λ = -2, and λ = 3 is the required value.

[JEE Main 2019 · 10 April · Shift II]

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