Q29 / 75 · JEE Main 2025 (2 Apr Shift 2) · Mathematics
If the system of equation
$\begin{aligned}<br/>& 2 x+\lambda y+3 z=5 \\
& 3 x+2 y-z=7 \\
& 4 x+5 y+\mu z=9<br/>\end{aligned}$
has infinitely many solutions, then $\left(\lambda^2+\mu^2\right)$ is equal to :
$\begin{aligned}<br/>& 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9<br/>\end{aligned}$
has infinitely many solutions, then $\left(\lambda^2+\mu^2\right)$ is equal to :
- A$22$
- B$18$
- Correct answerC$26$
- D$30$
Solution
$\Delta=0 \Rightarrow\left|\begin{array}{ccc}2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu\end{array}\right|=0$
$\Rightarrow 2(2 \mu+5)+\lambda(-4-3 \mu)+3(7)=0$
$\Rightarrow 4 \mu-3 \lambda \mu-4 \lambda+31=0$ ...(1)
$\begin{aligned}<br/>& \Delta_3=0 \Rightarrow\left|\begin{array}{lll}<br/>2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9<br/>\end{array}\right|=0 \\ & \Rightarrow 2(-17)+\lambda(1)+5(7)=0 \\ & \Rightarrow \lambda=-1<br/>\end{aligned}$
from equation (1)
$4 \mu+3 \mu+4+31=0 \Rightarrow \mu=-5$
$\therefore \lambda^2+\mu^2=26$
$\Rightarrow 2(2 \mu+5)+\lambda(-4-3 \mu)+3(7)=0$
$\Rightarrow 4 \mu-3 \lambda \mu-4 \lambda+31=0$ ...(1)
$\begin{aligned}<br/>& \Delta_3=0 \Rightarrow\left|\begin{array}{lll}<br/>2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9<br/>\end{array}\right|=0 \\ & \Rightarrow 2(-17)+\lambda(1)+5(7)=0 \\ & \Rightarrow \lambda=-1<br/>\end{aligned}$
from equation (1)
$4 \mu+3 \mu+4+31=0 \Rightarrow \mu=-5$
$\therefore \lambda^2+\mu^2=26$
Hence, the correct answer is option: 'C'.