JEE Main 2025 — Determinants Question with Solution
JEE Main 2025 (7 Apr Shift 2)
Question
Let the system of equations
$\begin{aligned}
& x+5 y-z=1 \\ & 4 x+3 y-3 z=7 \\ & 24 x+y+\lambda z=\mu
\end{aligned}\lambda, \mu \in \mathrm{R}x, y, z7 \leq x+y+z \leq 77$, is
$\begin{aligned}
& x+5 y-z=1 \\ & 4 x+3 y-3 z=7 \\ & 24 x+y+\lambda z=\mu
\end{aligned}\lambda, \mu \in \mathrm{R}x, y, z7 \leq x+y+z \leq 77$, is
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Show full solutionCorrect option: A
Correct answer
A3
Step-by-step explanation
For infinitely many solution
$\begin{aligned}
& \Delta=0 \\ & \left|\begin{array}{ccc}
1 & 5 & -1 \\ 4 & 3 & -3 \\ 24 & 1 & \lambda
\end{array}\right|=0 \\ & \Rightarrow 1(3 \lambda+3)-5(4 \lambda+72)-1(4-72)=0 \\ & \Rightarrow-17 \lambda+3-4 \times 72-4=0 \\ & \Rightarrow 17 \lambda=-289
\end{aligned}\begin{aligned} & \Delta 1=0 \\ & \Rightarrow\left|\begin{array}{ccc}1 & 5 & -1 \\ 7 & 3 & -3 \\ \mu & 1 & -17\end{array}\right|=0 \\ & \Rightarrow 1(-51+3)-5(-119+3 \mu)-1(7-3 \mu)=0 \\ & \Rightarrow-48+595-15 \mu-7+3 \mu=0 \\ & \Rightarrow 12 \mu=540\end{aligned}\begin{aligned} & x+5 y-z=1 \\ & 4 x+3 y-3 z=7 \\ & 24 x+y-17 z=45 \\ & \text { Let } z=1 \\ & x+5 y=1+\lambda] \times 4 \\ & 4 x+3 y=7+3 \lambda \\ & 4 x+20 y=4+4 \lambda \\ & \frac{-17 y=3-\lambda}{-17}\end{aligned}\begin{aligned} & \begin{aligned} \mathrm{y} & =\frac{\lambda-3}{17}, \mathrm{x}=1+\lambda-\frac{5 \lambda-15}{17} \\ & =\frac{32-12 \lambda}{17} \\ 7 & \leq \frac{\lambda-3}{17}+\frac{32+12 \lambda}{17}+\lambda \leq 77 \\ 7 & \leq \frac{30 \lambda+29}{17} \leq 77 \\ 3 & \leq \lambda \leq 42 \\ \lambda & =3,20,37\end{aligned}\end{aligned}$
$\begin{aligned}
& \Delta=0 \\ & \left|\begin{array}{ccc}
1 & 5 & -1 \\ 4 & 3 & -3 \\ 24 & 1 & \lambda
\end{array}\right|=0 \\ & \Rightarrow 1(3 \lambda+3)-5(4 \lambda+72)-1(4-72)=0 \\ & \Rightarrow-17 \lambda+3-4 \times 72-4=0 \\ & \Rightarrow 17 \lambda=-289
\end{aligned}\begin{aligned} & \Delta 1=0 \\ & \Rightarrow\left|\begin{array}{ccc}1 & 5 & -1 \\ 7 & 3 & -3 \\ \mu & 1 & -17\end{array}\right|=0 \\ & \Rightarrow 1(-51+3)-5(-119+3 \mu)-1(7-3 \mu)=0 \\ & \Rightarrow-48+595-15 \mu-7+3 \mu=0 \\ & \Rightarrow 12 \mu=540\end{aligned}\begin{aligned} & x+5 y-z=1 \\ & 4 x+3 y-3 z=7 \\ & 24 x+y-17 z=45 \\ & \text { Let } z=1 \\ & x+5 y=1+\lambda] \times 4 \\ & 4 x+3 y=7+3 \lambda \\ & 4 x+20 y=4+4 \lambda \\ & \frac{-17 y=3-\lambda}{-17}\end{aligned}\begin{aligned} & \begin{aligned} \mathrm{y} & =\frac{\lambda-3}{17}, \mathrm{x}=1+\lambda-\frac{5 \lambda-15}{17} \\ & =\frac{32-12 \lambda}{17} \\ 7 & \leq \frac{\lambda-3}{17}+\frac{32+12 \lambda}{17}+\lambda \leq 77 \\ 7 & \leq \frac{30 \lambda+29}{17} \leq 77 \\ 3 & \leq \lambda \leq 42 \\ \lambda & =3,20,37\end{aligned}\end{aligned}$
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