Let $S$ be the set of all real values of $k$ for which the system of linear equations $$ \begin{aligned}…
Let $S$ be the set of all real values of $k$ for which the system of linear equations
$$
\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
an empty set
equal to $\mathrm{R}-\{0\}$
equal to $\{0\}$
equal to $\mathrm{R}$
Solution
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, $k \in R-\{0\} \equiv S$