If the system of equations

11 x+y+lambda | vedclass

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Question

$ 11 x+y+\lambda z=-5 $

$ 2 x+3 y+5 z=3 $

$ 8 x-19 y-39 z=\mu$

for infinite sol.

$\mathrm{D}=\left|\begin{array}{ccc}11 & 1 & \lam

Vedclass · Accepted Answer

$47$

Vedclass · Answer

$ 11 x+y+\lambda z=-5 $

$ 2 x+3 y+5 z=3 $

$ 8 x-19 y-39 z=\mu$

for infinite sol.

$\mathrm{D}=\left|\begin{array}{ccc}11 & 1 & \lambda \ 2 & 3 & 5 \ 8 & -19 & -39\end{array}ight|=0$

$ \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$

$\mathrm{D}_1=\left|\begin{array}{ccc}-5 & 1 & -2 \ 3 & 3 & 5 \ \mu & -19 & -39\end{array}ight|=0$

$ \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$

If the system of equations

$ 11 x+y+\lambda z=-5 $

$ 2 x+3 y+5 z=3 $

$ 8 x-19 y-39 z=\mu$

has infinitely many solutions, then $\lambda^4-\mu$ is equal to :

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