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If the system of equations $x+k y-z=0$, $3 x-k y-z=0$ and $x-3 y+z=0$, has non-zero solution, then $\mathrm{k}$ is equal to
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The correct answer is:
1
The system has non- zero solution, if
$\begin{aligned}
&\left|\begin{array}{ccc}
1 & \mathrm{k} & -1 \\
3 & -\mathrm{k} & -1 \\
1 & -3 & 1
\end{array}\right|=0 \\
\Rightarrow & 1(-\mathrm{k}-3)-\mathrm{k}(3+1)-1(-9+\mathrm{k})=0 \\
\Rightarrow &-6 \mathrm{k}+6=0 \\
\Rightarrow & \mathrm{k}=1
\end{aligned}$
$\begin{aligned}
&\left|\begin{array}{ccc}
1 & \mathrm{k} & -1 \\
3 & -\mathrm{k} & -1 \\
1 & -3 & 1
\end{array}\right|=0 \\
\Rightarrow & 1(-\mathrm{k}-3)-\mathrm{k}(3+1)-1(-9+\mathrm{k})=0 \\
\Rightarrow &-6 \mathrm{k}+6=0 \\
\Rightarrow & \mathrm{k}=1
\end{aligned}$
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