Find out all solutions for the system Given the system
$$ \left[
\begin{array}{ccc|c}
  x_1&x_2&x_3&k\\
  x_1&x_2&kx_3&1\\
  x_1&kx_2&x_3&1\\
  kx_1&x_2&x_3&1\\
\end{array}
\right] $$
I tried to solve this...It looks simple but I found a problem at the end...
$$ \left[
\begin{array}{ccc|c}
  1&1&1&k\\
  1&1&k&1\\
  1&k&1&1\\
  k&1&1&1\\
\end{array}
\right] $$
$$ \left[
\begin{array}{ccc|c}
  1&1&1&k\\
  0&0&k-1&1-k\\
  0&k-1&0&1-k\\
  k-1&0&0&1-k\\
\end{array}
\right] $$
$$ \left[
\begin{array}{ccc|c}
  1&1&1&k\\
  0&0&1& \frac{1-k}{k-1}\\
  0&1&0& \frac{1-k}{k-1}\\
  1&0&0& \frac{1-k}{k-1}\\
\end{array}
\right] $$
Finally,
$$ \left[
\begin{array}{ccc|c}
  1&0&0& -1\\
  0&1&0& -1\\
  0&0&1& -1\\
  0&0&0&k+3\\
\end{array}
\right] $$
There is no way to get infinitely many solutions.
$$$$
However, I tried to put the system into other online calculator...
Click me$$$$
There is infinitely many solutions when k=4 or other numbers. What's wrong of my work...

The system has solutions only if 
$\det(A|B)=0$, otherwise $\text{rank }(A|B)=4>\text{rank }(A)$
that is $k^4-6 k^2+8 k-3=0\to (k-1)^3 (k+3)=0$ 
for $k=-3$ and $k=1$
for $k=-3$ we get the system
$\begin {cases}
x+y+z=-3\\
x+y-3 z=1\\
x-3 y+z=1\\
-3x+y+z=1\\
\end{cases}
$
which has solution $(-1,-1,-1)$
for $k=1$ we get
$\begin {cases}
x+y+z=1\\
x+y+ z=1\\
x+y+z=1\\
x+y+z=1\\
\end{cases}
$
that is 
$x+y+z=1$ which gives infinite solutions
$(t,u,1-t-u)$
Hope this helps
