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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 :
Options:
$\begin{array}{r}
11 x+y+\lambda z=-5 \\
2 x+3 y+5 z=3 \\
8 x-19 y-39 z=\mu
\end{array}$
has infinitely many solutions, then $\lambda^4-\mu$ is equal to :
Solution:
2518 Upvotes
Verified Answer
The correct answer is:
47
$\begin{aligned}
& 11 x+y+\lambda z=-5 \\
& 2 x+3 y+5 z=3 \\
& 8 x-19 y-39 z=\mu
\end{aligned}$
for infinite sol.
$\begin{aligned}
& \mathrm{D}=\left|\begin{array}{ccc}
11 & 1 & \lambda \\
2 & 3 & 5 \\
8 & -19 & -39
\end{array}\right|=0 \\
& \Rightarrow 11(-117+95)-1(-78-40)+\lambda(-38-24) \\
& \Rightarrow 11(-22)+118-\lambda(62)=0 \\
& \Rightarrow 62 \lambda=118-242 \\
& \Rightarrow \lambda=\frac{-124}{62}=-2
\end{aligned}$
$\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc}-5 & 1 & -2 \\ 3 & 3 & 5 \\ \mu & -19 & -39\end{array}\right|=0 \\ & \Rightarrow-5(-117+95)-1(-117-5 \mu)-2(-57-3 \mu)=0 \\ & \Rightarrow-5(-22)+117+5 \mu+114+6 \mu=0 \\ & \Rightarrow 11 \mu=-110-231=-341 \\ & \Rightarrow \mu=-31 \\ & \lambda^4-\mu=(-2)^4-(-31)=16+31=47\end{aligned}$
& 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.
$\begin{aligned}
& \mathrm{D}=\left|\begin{array}{ccc}
11 & 1 & \lambda \\
2 & 3 & 5 \\
8 & -19 & -39
\end{array}\right|=0 \\
& \Rightarrow 11(-117+95)-1(-78-40)+\lambda(-38-24) \\
& \Rightarrow 11(-22)+118-\lambda(62)=0 \\
& \Rightarrow 62 \lambda=118-242 \\
& \Rightarrow \lambda=\frac{-124}{62}=-2
\end{aligned}$
$\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc}-5 & 1 & -2 \\ 3 & 3 & 5 \\ \mu & -19 & -39\end{array}\right|=0 \\ & \Rightarrow-5(-117+95)-1(-117-5 \mu)-2(-57-3 \mu)=0 \\ & \Rightarrow-5(-22)+117+5 \mu+114+6 \mu=0 \\ & \Rightarrow 11 \mu=-110-231=-341 \\ & \Rightarrow \mu=-31 \\ & \lambda^4-\mu=(-2)^4-(-31)=16+31=47\end{aligned}$
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