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...

Let us add the last three equations to gather to get: $$(k+2)(x_1+x_2+x_3) =3$$
Then we use the first one to get, $$(k+2)(k)=3 \implies k=1, -3$$
So, ?
