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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Determinants

If the system of equation aligned & 2 x+λ y+3 z=5 & 3 x+2 y-z=7 & 4...

If the system of equation
<br/>2x+λy+3z=53x+2y−z=74x+5y+μz=9<br/>\begin{aligned}<br/>& 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9<br/>\end{aligned}
has infinitely many solutions, then (λ2+μ2)\left(\lambda^2+\mu^2\right) is equal to :

Options

  1. A2222
  2. B1818
  3. C2626
  4. D3030

Correct answer

C. 2626

Step-by-step solution

\Delta=0 \Rightarrow\left|2λ332−145μ\begin{array}{ccc}2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu\end{array}\right|=0
⇒2(2μ+5)+λ(−4−3μ)+3(7)=0\Rightarrow 2(2 \mu+5)+\lambda(-4-3 \mu)+3(7)=0
⇒4μ−3λμ−4λ+31=0\Rightarrow 4 \mu-3 \lambda \mu-4 \lambda+31=0 ...(1)
<br/>Δ3=0⇒∣<br/>2λ5327459<br/>∣=0⇒2(−17)+λ(1)+5(7)=0⇒λ=−1<br/>\begin{aligned}<br/>& \Delta_3=0 \Rightarrow\left|\begin{array}{lll}<br/>2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9<br/>\end{array}\right|=0 \\ & \Rightarrow 2(-17)+\lambda(1)+5(7)=0 \\ & \Rightarrow \lambda=-1<br/>\end{aligned}
from equation (1)
4μ+3μ+4+31=0⇒μ=−54 \mu+3 \mu+4+31=0 \Rightarrow \mu=-5
∴λ2+μ2=26\therefore \lambda^2+\mu^2=26
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