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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Determinants

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

If the system of equations aligned & 2 x-y+z=4 & 5 x+λ y+3 z=12 &...

If the system of equations 2x−y+z=45x+λy+3z=12100x−47y+μz=212\begin{aligned} & 2 x-y+z=4 \\ & 5 x+\lambda y+3 z=12 \\ & 100 x-47 y+\mu z=212 \end{aligned} has infinitely many solutions, then μ−2λ\mu-2 \lambda is equal to

Options

  1. A5757
  2. B5959
  3. C5555
  4. D5656

Correct answer

A. 5757

Step-by-step solution

2x−y+z=45x+λy+3z=12100x−47y+μz=212Δx=Δy=Δz=0Δz=∣2−145λ12100−47212∣=0⇒2(212λ+564)+1(1060−1200)⇒424λ+1128−140−940−400λ=0⇒λ=−2Δy=∣2−115−23100−47μ∣=0⇒2(−2μ+141)+1(5μ−300)+1(−235+200)=0⇒μ=53μ−2λ=53−2(−2)=57\begin{aligned} & 2 x-y+z=4 \\ & 5 x+\lambda y+3 z=12 \\ & 100 x-47 y+\mu z=212 \\ & \Delta x=\Delta y=\Delta z=0 \\ & \Delta z=\left|\begin{array}{ccc}2 & -1 & 4 \\ 5 & \lambda & 12 \\ 100 & -47 & 212\end{array}\right|=0 \\ & \Rightarrow 2(212 \lambda+564)+1(1060-1200) \\ & \Rightarrow 424 \lambda+1128-140-940-400 \lambda=0 \\ & \Rightarrow \lambda=-2 \\ & \quad \Delta y=\left|\begin{array}{ccc}2 & -1 & 1 \\ 5 & -2 & 3 \\ 100 & -47 & \mu\end{array}\right|=0 \\ & \Rightarrow 2(-2 \mu+141)+1(5 \mu-300)+1(-235+200)=0 \\ & \Rightarrow \quad \mu=53 \\ & \mu-2 \lambda=53-2(-2)=57\end{aligned}
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