Answer
The given system has a unique solution if and only if $k \ne \pm 2$
Work Step by Step
$A=\begin{bmatrix}
1 &k\\
k &4
\end{bmatrix}$
$x=\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}$
$C=\begin{bmatrix}
b_1\\
b_2
\end{bmatrix}$
$\rightarrow \det A=4-k^2$
The given system has a unique solution if and only if
$\det A\ne0$
$4-k^2 \ne 0$
Hence, $k \ne \pm 2$