3x3 system of equations I have the following system of equations:
\begin{equation}
x+y+kz = 1
\end{equation}
\begin{equation}
x+ky+z = 1
\end{equation}
\begin{equation}
kx+y+z =1
\end{equation}
where $x,y,z \in \mathbf{R}$.
For what values of $k \in \mathbf{R} $ 
i) the system has a single solution 
ii) the system has multiple solutions
iii) the system has no solution
Can anyone help with that please?
I wrote the above system in a matrix form: 
\begin{equation*}
\begin{array}{rrr|r}
1 & 1 & k & 1 \\
1 & k & 1 & 1\\
k & 1 & 1 & 1
 \end{array}
\end{equation*}
I am not sure how to proceed.. If I subtract the second row from the first (1)-(2) I have the following:
\begin{equation*}
\begin{array}{rrr|r}
1 & 1 & k & 1 \\
0 & 1-k & k-1 & 0\\
k & 1 & 1 & 1
 \end{array}
\end{equation*}
Then if I subtract the third row from the first row (1)-(3):
\begin{equation*}
\begin{array}{rrr|r}
1 & 1 & k & 1 \\
0 & 1-k & k-1 & 0\\
1-k & 0 & k-1 & 0
 \end{array}
\end{equation*}

$$\Delta=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ k & 1 & 1 \end{array}\right)\right|=-k^3+3k-2=-(k+2)(k-1)^2,$$
$$\Delta_x=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ 1 & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$
$$\Delta_y=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & 1 & 1\\ k & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$
$$\Delta_z=\left|\left(\begin{array}{cc}1 & 1 & 1\\1 & k & 1\\ k & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$ which gives the answer:
i) $k\neq-2$ and $k\neq1$;
ii) $k=1$;
iii) $k=-2$.
