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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Determinants

If the system of equations aligned & x+2 y-3 z=2 & 2 x+λ y+5 z=5 &...

If the system of equations x+2y−3z=22x+λy+5z=514x+3y+μz=33\begin{aligned} & x+2 y-3 z=2 \\ & 2 x+\lambda y+5 z=5 \\ & 14 x+3 y+\mu z=33 \end{aligned} has infinitely many solutions, then λ+μ\lambda+\mu is equal to :

Options

  1. A13
  2. B10
  3. C12
  4. D11

Correct answer

C. 12

Step-by-step solution

Δ=∣12−32λ5143μ∣=0⇒λμ+42λ−4μ+107=0Δ1=∣22−35λ5333μ∣=0⇒2λμ+99λ−10μ+255=0Δ2=∣12−32551433μ∣=0⇒μ=13\begin{aligned} & \Delta=\left|\begin{array}{ccc} 1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu \end{array}\right|=0 \Rightarrow \lambda \mu+42 \lambda-4 \mu+107=0 \\ & \Delta_1=\left|\begin{array}{ccc} 2 & 2 & -3 \\ 5 & \lambda & 5 \\ 33 & 3 & \mu \end{array}\right|=0 \Rightarrow 2 \lambda \mu+99 \lambda-10 \mu+255=0 \\ & \Delta_2=\left|\begin{array}{ccc} 1 & 2 & -3 \\ 2 & 5 & 5 \\ 14 & 33 & \mu \end{array}\right|=0 \Rightarrow \mu=13 \end{aligned} Also, λ=−1\lambda=-1 Hence, λ+μ=13−1=12\lambda+\mu=13-1=12
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