Answer
The given system has a unique solution if and only if $k \ne 1$ or $k \ne0$
Work Step by Step
$A=\begin{bmatrix}
1 &k & 0|2\\
k &1&1|1\\
1 & 1 & 1 | 1
\end{bmatrix}$
The determinant for $\begin{bmatrix}
1 &k & 0\\
k &1&1\\
1 & 1 & 1
\end{bmatrix}$ is:
$\rightarrow \det A=k-k^2=k(1-k)$
The given system has a unique solution if and only if
$\det A\ne0$
$k(1-k) \ne 0$
Hence, $k \ne 1$ or $k \ne0$
