If the system of equations $\begin{array}{r} 11 x+y+\lambda z=-5 \ 2 x+3 y+5 z=3 \ 8 x-19 y-39 z=\mu \end{array}$ has infinitely many solutions, then $\lambda^4-\mu$ is equal to :

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Aicharya · Accepted Answer

$\begin{aligned} & 11 x+y+\lambda z=-5 \ & 2 x+3 y+5 z=3 \ & 8 x-19 y-39 z=\mu \end{aligned}$ for infinite sol. $\begin{aligned} & \mathrm{D}=\left|\begin{array}{ccc} 11 & 1 & \lambda \ 2 & 3 & 5 \ 8 & -19 & -39 \end{array}ight|=0 \ & \Rightarrow 11(-117+95)-1(-78-40)+\lambda(-38-24) \ & \Rightarrow 11(-22)+118-\lambda(62)=0 \ & \Rightarrow 62 \lambda=118-242 \ & \Rightarrow \lambda=\frac{-124}{62}=-2 \end{aligned}$ $\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc}-5 & 1 & -2 \ 3 & 3 & 5 \ \mu & -19 & -39\end{array}ight|=0 \ & \Rightarrow-5(-117+95)-1(-117-5 \mu)-2(-57-3 \mu)=0 \ & \Rightarrow-5(-22)+117+5 \mu+114+6 \mu=0 \ & \Rightarrow 11 \mu=-110-231=-341 \ & \Rightarrow \mu=-31 \ & \lambda^4-\mu=(-2)^4-(-31)=16+31=47\end{aligned}$

If the system of equations $\begin{array}{r} 11 x+y+\lambda z=-5 \\ 2 x+3 y+5 z=3 \\ 8 x-19 y-39 z=\mu…

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