Q29 / 75 · JEE Main 2025 (24 Jan Shift 1) · Mathematics

If the system of equations $\begin{aligned} & 2 x-y+z=4 \\ & 5 x+\lambda y+3 z=12 \\ & 100 x-47 y+\mu z=212 \end{aligned}$ has infinitely many solutions, then $\mu-2 \lambda$ is equal to

  • Correct answer
    A
    $57$
  • B
    $59$
  • C
    $55$
  • D
    $56$

Solution

$\begin{aligned} & 2 x-y+z=4 \\ & 5 x+\lambda y+3 z=12 \\ & 100 x-47 y+\mu z=212 \\ & \Delta x=\Delta y=\Delta z=0 \\ & \Delta z=\left|\begin{array}{ccc}2 & -1 & 4 \\ 5 & \lambda & 12 \\ 100 & -47 & 212\end{array}\right|=0 \\ & \Rightarrow 2(212 \lambda+564)+1(1060-1200) \\ & \Rightarrow 424 \lambda+1128-140-940-400 \lambda=0 \\ & \Rightarrow \lambda=-2 \\ & \quad \Delta y=\left|\begin{array}{ccc}2 & -1 & 1 \\ 5 & -2 & 3 \\ 100 & -47 & \mu\end{array}\right|=0 \\ & \Rightarrow 2(-2 \mu+141)+1(5 \mu-300)+1(-235+200)=0 \\ & \Rightarrow \quad \mu=53 \\ & \mu-2 \lambda=53-2(-2)=57\end{aligned}$

Hence, the correct answer is option: 'A'.

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