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The system of equations $$ \begin{array}{l} \lambda x+y+3 z=0 \\ 2 x+\mu y-z=0 \\ 5 x+7 y+z=0 \end{array} $$ has infinitely many solutions in $R$. Then,
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Verified Answer
The correct answer is:
$\lambda=1, \mu=3$
Given, system of equations
$$
\begin{array}{l}
\lambda x+y+3 z=0 \\
2 x+\mu y-z=0 \\
5 x+7 y+z=0
\end{array}
$$
System has infinitely many solutions in $R$, if
$$
\left|\begin{array}{ccc}
\lambda & 1 & 3 \\
2 & \mu & -1 \\
5 & 7 & 1
\end{array}\right|=0
$$
$\Rightarrow \quad \lambda(\mu+7)-1(2+5)+3(14-5 \mu)=0$
$\Rightarrow \quad \lambda(\mu+7)-7+3(14-5 \mu)=0$
$\Rightarrow \quad \lambda \mu+7 \lambda-7+42-15 \mu=0$
$\Rightarrow \quad \lambda \mu+7 \lambda-15 \mu+35=0$
By checking the options, we get option (c) $(\lambda=1, \mu=3)$ satisfies the given equation.
$$
\begin{array}{l}
\lambda x+y+3 z=0 \\
2 x+\mu y-z=0 \\
5 x+7 y+z=0
\end{array}
$$
System has infinitely many solutions in $R$, if
$$
\left|\begin{array}{ccc}
\lambda & 1 & 3 \\
2 & \mu & -1 \\
5 & 7 & 1
\end{array}\right|=0
$$
$\Rightarrow \quad \lambda(\mu+7)-1(2+5)+3(14-5 \mu)=0$
$\Rightarrow \quad \lambda(\mu+7)-7+3(14-5 \mu)=0$
$\Rightarrow \quad \lambda \mu+7 \lambda-7+42-15 \mu=0$
$\Rightarrow \quad \lambda \mu+7 \lambda-15 \mu+35=0$
By checking the options, we get option (c) $(\lambda=1, \mu=3)$ satisfies the given equation.
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