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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Determinants

If the system of equations aligned & (λ-1) x+(λ-4) y+λ z=5 & λ...

If the system of equations (λ−1)x+(λ−4)y+λz=5λx+(λ−1)y+(λ−4)z=7(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

Options

  1. A66
  2. B1010
  3. C2020
  4. D1212

Correct answer

D. 1212

Step-by-step solution

(λ−1)x+(λ−4)y+λz=5λx+(λ−1)y+(λ−4)z=7(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} For infinitely many solutions D=∣λ−1λ−4λλλ−1λ−4λ+1λ+2−(λ+2)∣=0(λ−3)(2λ+1)=0Dx=∣5λ−4λ7λ−1λ−49λ+2−(λ+2)∣=02(3−λ)(23−2λ)=0λ=3∴λ2+λ=9+3=12\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}
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