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Question:
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Let $S$ be the set of all real values of $k$ for which the system of linear equations
$$
\begin{aligned}
&x+y+z=2 \\
&2 x+y-z=3 \\
&3 x+2 y+k z=4
\end{aligned}
$$
has a unique solution. Then $S$ is
Options:
$$
\begin{aligned}
&x+y+z=2 \\
&2 x+y-z=3 \\
&3 x+2 y+k z=4
\end{aligned}
$$
has a unique solution. Then $S$ is
Solution:
2683 Upvotes
Verified Answer
The correct answer is:
equal to $\mathrm{R}-\{0\}$
equal to $\mathrm{R}-\{0\}$
The system of linear equations is:
$$
\begin{aligned}
&x+y+z=2 \\
&2 x+y-z=3 \\
&3 x+2 y+k z=4
\end{aligned}
$$
As, system has unique solution.
$$
\begin{aligned}
&\text { So, }\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 1 & -1 \\
3 & 2 & k
\end{array}\right| \neq 0 \\
&\Rightarrow k+2-(2 k+3)+1 \neq 0 \\
&\Rightarrow k \neq 0
\end{aligned}
$$
Hence, $k \in R-\{0\} \equiv S$
$$
\begin{aligned}
&x+y+z=2 \\
&2 x+y-z=3 \\
&3 x+2 y+k z=4
\end{aligned}
$$
As, system has unique solution.
$$
\begin{aligned}
&\text { So, }\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 1 & -1 \\
3 & 2 & k
\end{array}\right| \neq 0 \\
&\Rightarrow k+2-(2 k+3)+1 \neq 0 \\
&\Rightarrow k \neq 0
\end{aligned}
$$
Hence, $k \in R-\{0\} \equiv S$
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