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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Determinants

If the system of equations x+4 y-z=λ, 7 x+9 y+μ z=-3,5 x+y+2 z=-1...

If the system of equations x+4y−z=λ,7x+9y+μz=−3,5x+y+2z=−1x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1 has infinitely many solutions, then (2μ+3λ)(2 \mu+3 \lambda) is equal to :

Options

  1. A3
  2. B-3
  3. C-2
  4. D2

Correct answer

B. -3

Step-by-step solution

Δ=∣14−179μ512∣=0⇒(18−μ)−4(14−5μ)−(7−45)=0⇒μ=0Δ=Δx=Δy=Δz=0 (For infinite solution) Δx=∣λ4−1−39μ−112∣=0λ(18−μ)−4(−6+μ)−1(−3+9)=018λ+24−6=0⇒λ=−1\begin{aligned} & \Delta=\left|\begin{array}{ccc}1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2\end{array}\right|=0 \\ & \Rightarrow(18-\mu)-4(14-5 \mu)-(7-45)=0 \Rightarrow \mu=0 \\ & \Delta=\Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0 \text{ (For infinite solution) } \\ & \Delta_{\mathrm{x}}=\left|\begin{array}{ccc}\lambda & 4 & -1 \\ -3 & 9 & \mu \\ -1 & 1 & 2\end{array}\right|=0 \\ & \lambda(18-\mu)-4(-6+\mu)-1(-3+9)=0 \\ & 18 \lambda+24-6=0 \Rightarrow \lambda=-1\end{aligned}
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