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QJEEM25Madd2b: JEE Main 2025 Mathematics - Class 12 JEEM25Madd2b Solution:   If the system of equations has infinitely many…

 

If the system of equations

amp;(λ−1)x+(λ−4)y+λz=5amp;λx+(λ−1)y+(λ−4)z=7amp;(λ+1)x+(λ+2)y−(λ+2)z=9\begin{aligned}& (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\& \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\& (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9\end{aligned}

has infinitely many solutions, then λ2+λ\lambda^2+\lambda is equal to

Multiple Choice Options

  • A 20
  • B 10
  • C 6
  • D 12

Step-by-Step Verified Solution

amp;amp;(λ−1)x+(λ−4)y+λz=5amp;λx+(λ−1)y+(λ−4)z=7amp;(λ+1)x+(λ+2)y−(λ+2)z=9amp; For infinitely many solutions amp;amp;D=∣λ−1amp;λ−4amp;λλamp;λ−1amp;λ−4λ+1amp;λ+2amp;−(λ+2)∣=0amp;(λ−3)(2λ+1)=0amp;Dx=∣5amp;λ−4amp;λ7amp;λ−1amp;λ−49amp;λ+2amp;−(λ+2)∣=0amp;2(3−λ)(23−2λ)=0amp;λ=3amp;∴λ2+λ=9+3=12\begin{aligned}&\begin{aligned}& (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\& \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\& (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9\end{aligned}\\&\text { For infinitely many solutions }\\&\begin{aligned}& \mathrm{D}=\left|\begin{array}{ccc}\lambda-1 & \lambda-4 & \lambda \\\lambda & \lambda-1 & \lambda-4 \\\lambda+1 & \lambda+2 & -(\lambda+2)\end{array}\right|=0 \\& (\lambda-3)(2 \lambda+1)=0 \\& \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc}5 & \lambda-4 & \lambda \\7 & \lambda-1 & \lambda-4 \\9 & \lambda+2 & -(\lambda+2)\end{array}\right|=0 \\& 2(3-\lambda)(23-2 \lambda)=0 \\& \lambda=3 \\& \therefore \lambda^2+\lambda=9+3=12\end{aligned}\end{aligned}

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