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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Determinants

If the system of equations array r 11 x+y+λ z=-5 2 x+3 y+5 z=3 8...

If the system of equations 11x+y+λz=−52x+3y+5z=38x−19y−39z=μ\begin{array}{r} 11 x+y+\lambda z=-5 \\ 2 x+3 y+5 z=3 \\ 8 x-19 y-39 z=\mu \end{array} has infinitely many solutions, then λ4−μ\lambda^4-\mu is equal to :

Options

  1. A51
  2. B45
  3. C47
  4. D49

Correct answer

C. 47

Step-by-step solution

11x+y+λz=−52x+3y+5z=38x−19y−39z=μ\begin{aligned} & 11 x+y+\lambda z=-5 \\ & 2 x+3 y+5 z=3 \\ & 8 x-19 y-39 z=\mu \end{aligned} for infinite sol. D=∣111λ2358−19−39∣=0⇒11(−117+95)−1(−78−40)+λ(−38−24)⇒11(−22)+118−λ(62)=0⇒62λ=118−242⇒λ=−12462=−2\begin{aligned} & \mathrm{D}=\left|\begin{array}{ccc} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{array}\right|=0 \\ & \Rightarrow 11(-117+95)-1(-78-40)+\lambda(-38-24) \\ & \Rightarrow 11(-22)+118-\lambda(62)=0 \\ & \Rightarrow 62 \lambda=118-242 \\ & \Rightarrow \lambda=\frac{-124}{62}=-2 \end{aligned}D1=∣−51−2335μ−19−39∣=0⇒−5(−117+95)−1(−117−5μ)−2(−57−3μ)=0⇒−5(−22)+117+5μ+114+6μ=0⇒11μ=−110−231=−341⇒μ=−31λ4−μ=(−2)4−(−31)=16+31=47\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc}-5 & 1 & -2 \\ 3 & 3 & 5 \\ \mu & -19 & -39\end{array}\right|=0 \\ & \Rightarrow-5(-117+95)-1(-117-5 \mu)-2(-57-3 \mu)=0 \\ & \Rightarrow-5(-22)+117+5 \mu+114+6 \mu=0 \\ & \Rightarrow 11 \mu=-110-231=-341 \\ & \Rightarrow \mu=-31 \\ & \lambda^4-\mu=(-2)^4-(-31)=16+31=47\end{aligned}
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