Find the value of k, (if any), for which the system below has unique, infinite or no solution. The system of equations are:
$\begin{cases}x+y+kz = 1\\x+ky+z=1\\kx+y+z=1\\
\end{cases}$
I am looking to finding values of $k$, for which this system has either no solutions, infinite many solutions or a unique solution (if any).
I've had a look around the web for some help, and know that this question has been asked, but in a way that I have not yet learnt (something to do with determinants I think).
However, I have only been working with reducing the matrix to REF/RREF and thus I'm trying to find rows, such as $( 0 \ 0 \ 0 \mid 2)$ and $( 0 \ 0 \ 0 \mid 0)$ etc to show if we have no solutions or infinite many solutions, for instance.
So far, I've got the following:
$\left[\begin{array}{ccc|c}1&1&k&1\\1&k&1&1\\k&1&1&1\end{array}\right]$
Then, $R_2$$\mapsto$$R_2-R_1$
$\left[\begin{array}{ccc|c}1&1&k&1\\0&k-1&1-k&0\\k&1&1&1\end{array}\right]$
Then, $R_2$ $\leftrightarrow$ $R_3$ 
$\left[\begin{array}{ccc|c}1&1&k&1\\k&1&1&1\\0&k-1&1-k&0\end{array}\right]$
This is where I got stuck, and I'm not sure how to continue.
Thank you in advance, and I'm sorry if this question has been asked and solved in the same way as presented above.
Cheers.
 A: As I mentioned in a comment, there is at least one another post about the same problem: System of Linear Equations - how many solutions?
Let me address your attempt. (I think that as soon as you are satisfied with the feedback you got on your attempt to solve the problem, this post might be closed as a duplicate of some other post asking the same question.)

$\left[\begin{array}{ccc|c}1&1&k&1\\k&1&1&1\\0&k-1&1-k&0\end{array}\right]$
This is where I got stuck, and I'm not sure how to continue.

Now a reasonable step would be to divide the last row by $(k-1)$. But this is only possible if $k-1\ne0$.
So you can:

*

*Solve the case $k=1$ separately.

*For $k\ne 1$ divide the last row by $(k-1)$ which leads you to a simpler matrix - you should try whether you are able to continue from there.

$$\left[\begin{array}{ccc|c}1&1&k&1\\k&1&1&1\\0&1&-1&0\end{array}\right]$$
A: I would restart from:
$$\left[\begin{array}{ccc|c}1&1&k&1\\0&k-1&1-k&0\\k&1&1&1\end{array}\right]$$
then $\,R_3\leftarrow R_3-kR_1$:
$$\left[\begin{array}{ccc|c}1&1&k&1\\0&k-1&1-k&0\\0&1-k&1-k^2&1-k\end{array}\right]$$
$R_3\leftarrow R_3+R_2$:
$$\left[\begin{array}{ccc|c}1&1&k&1\\0&k-1&1-k&0\\0&0&(1-k)(2+k)&1-k\end{array}\right]\tag{1}$$
Suppose for now $\boldsymbol {k\neq 1,-2}.$
$R_2\leftarrow \frac{1}{k-1}R_2$:
$$\left[\begin{array}{ccc|c}1&1&k&1\\0&1&-1&0\\0&0&(1-k)(2+k)&1-k\end{array}\right]$$
then  $\,R_2\leftarrow R_2+\frac1{(1-k)(2+k)}R_3$ and $\,R_1\leftarrow R_1-\frac k{(1-k)(2+k)}R_3$:
$$\left[\begin{array}{ccc|c}1&1&0&\dfrac2{2+k}\\0&1&0&\dfrac1{2+k}\\0&0&(1-k)(2+k)&1-k\end{array}\right]$$
and finally $\,R_1\leftarrow R_1-R_2$, $\,R_3\leftarrow \frac1{(1-k)(2+k)}R_3$
$$\left[\begin{array}{ccc|c}1&0&0&\dfrac1{2+k}\\0&1&0&\dfrac1{2+k}\\0&0&1&\dfrac1{2+k}\end{array}\right]$$
The solution is $\enspace x=y=z=\dfrac1{2+k}$.
If $\boldsymbol{k=1}$: the matrix (1) becomes
$$\left[\begin{array}{ccc|c}1&1&1&1\\0&0&0&0\\0&0&0&0\end{array}\right]$$
so the solutions consists in the plane $\,x+y+z=1$.
If $\boldsymbol{k=-2}$: the matrix becomes:
$$\left[\begin{array}{ccc|c}1&1&-2&1\\0&1&-1&0\\0&0&0&-1\end{array}\right]$$
and the last row shows there's no solution.
