Q49 / 90 · JEE Main 2024 (09 Apr Shift 1) · Mathematics

Let $\lambda, \mu \in \mathbf{R}$. If the system of equations $\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned}$ has infinitely many solutions, then $\mu+2 \lambda$ is equal to :

  • A
    24
  • Correct answer
    B
    25
  • C
    22
  • D
    27

Solution

$\begin{array}{l} 3 x+5 y+\lambda z=3 \\ 7 x+11 y-9 z=2 \\ 97 x+155 y-189 z=\mu \\ 93 x+155 y+31 \lambda z=93 \\ 97 x+155 y-189 z=\mu \\ -\quad+\quad+\quad- \\ \hline-4 x+(31 \lambda+189) z=93-\mu \\ 1085 x+1705 y-1395 z=310 \\ 1067 x+1705 y-2079 z=11 \mu \\ -\quad+\quad-\quad \\ \hline 18 x+684 z=310-11 \mu \end{array}$ $\begin{aligned} & -36 x+9(31 \lambda+189) z=9(93-\mu) \\ & 36 x+1368 z=2(310-11 \mu) \\ & (279 \lambda+3069) z=1457-31 \mu \end{aligned}$ for infinite solutions - $\begin{aligned} & \lambda=\frac{-3069}{279}=\frac{-341}{31} \\ & \mu=\frac{1457}{31} \end{aligned}$ $\mu+2 \lambda=\frac{1457-682}{31}=\frac{775}{31}=25$

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

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