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QJEEM24Ma2667: JEE Main 2024 Mathematics - Class 12 JEEM24Ma2667 Solution: If the system of equations has infinitely many solutions, then…

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 :

Multiple Choice Options

  • A 51
  • B 45
  • C 47
  • D 49

Step-by-Step Verified Solution

amp;11x+y+λz=−5amp;2x+3y+5z=3amp;8x−19y−39z=μamp;Δ=0⇒∣11amp;1amp;λ2amp;3amp;58amp;−19amp;−39∣=0amp;11(−39.3+19.5)−1(−39.2−40)+λ(−38−24)=0amp;=11(−117+95)−1(−118)−62λ=0amp;=−242+118=62λamp;⇒λ=−2amp;Δ2=0amp;⇒∣11amp;1amp;−52amp;3amp;38amp;−19amp;μ∣=0\begin{aligned}& 11 x+y+\lambda z=-5 \\& 2 x+3 y+5 z=3 \\& 8 x-19 y-39 z=\mu \\& \Delta=0 \Rightarrow\left|\begin{array}{ccc}11 & 1 & \lambda \\2 & 3 & 5 \\8 & -19 & -39\end{array}\right|=0 \\& 11(-39.3+19.5)-1(-39.2-40)+\lambda(-38-24)=0 \\& =11(-117+95)-1(-118)-62 \lambda=0 \\& =-242+118=62 \lambda \\& \Rightarrow \lambda=-2 \\& \Delta 2=0 \\& \Rightarrow\left|\begin{array}{ccc}11 & 1 & -5 \\2 & 3 & 3 \\8 & -19 & \mu\end{array}\right|=0\end{aligned}

amp;11(3μ+57)−1(2μ−24)−5(−38−24)=0amp;33μ+627−2μ+24+310=0amp;μ=−31amp;⇒λ4−31amp;=16+31amp;=47\begin{aligned}& 11(3 \mu+57)-1(2 \mu-24)-5(-38-24)=0 \\& 33 \mu+627-2 \mu+24+310=0 \\& \mu=-31 \\& \Rightarrow \lambda^4-31 \\& =16+31 \\& =47\end{aligned}

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