Answer
$k\neq -1$
Work Step by Step
The augmented matrix is :
$\begin{array}{l}
\left[ {\begin{array}{*{20}{c}}
1&{ - 1}&2&0\\
{ - 1}&1&{ - 1}&0\\
1&k&1&0
\end{array}} \right]\\
{R_2} \to {R_2} + {R_1},{R_3} \to {R_3} - {R_1}\\
\left[ {\begin{array}{*{20}{c}}
1&{ - 1}&2&0\\
0&0&1&0\\
0&{k + 1}&{ - 1}&0
\end{array}} \right]\\
{R_1} \to {R_1} - 2{R_2};{R_3} \to {R_3} + {R_2}\\
\left[ {\begin{array}{*{20}{c}}
1&{ - 1}&0&0\\
0&0&1&0\\
0&{k + 1}&0&0
\end{array}} \right]\\
{R_3} \leftrightarrow {R_2}\\
\left[ {\begin{array}{*{20}{c}}
1&{ - 1}&0&0\\
0&{k + 1}&0&0\\
0&0&1&0
\end{array}} \right]
\end{array}$
This corresponds to the system
$x-y=0\\
(k+1)y=0\\
z=0$
If $k=-1$, then equation (2) becomes $0y=0$, which corresponds to the zero row, implying $y$ is a free variable.
Hence, if $k\ne -1$, then $x=y=z=0$ and the system has exactly one solution.
