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 :

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Aicharya · Accepted Answer

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

Let $\lambda, \mu \in \mathbf{R}$. If the system of equations $\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7…

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