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*}
 A: I think Michael Rozenberg’s suggestion to use Cramer’s rule is a good one because the determinants are all very easy to compute. However, if you must proceed via row-reduction, you need to proceed systematically: your goal should be to end up with all zeros below each pivot. You also need to be a bit careful about the operations that you perform. The operation $R_n\to cR_n+R_m$, which is what you did for your second step, isn’t an elementary row operation, and doing thing like this can cause problems for you in other calculations, such as computing the determinant of a matrix. This is really a combination of two elementary operations: multiply a row by a constant and add one row to another.  
Anyway, with the goal of zeroing out everything below the $1$ in the upper-left corner, you should add $-k$ times the first row the the third to obtain $$\left[\begin{array}{ccc|c}1&1&k&1\\0&k-1&1-k&0\\0&1-k&1-k^2&1-k\end{array}\right].$$ At this point, it should be obvious that when $k=1$, the last two rows are both zero, so there’s an infinite number of solutions in that case. We continue with the row reduction by adding the second row to the third: $$\left[\begin{array}{ccc|c}1&1&k&1\\0&k-1&1-k&0\\0&0&2-k-k^2&1-k\end{array}\right].$$ Now examine the last row to determine which values of $k$ result in none, one or an infinite number of solutions.
A: $$\Delta=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ k & 1 & 1 \end{array}\right)\right|=-k^3+3k-2=-(k+2)(k-1)^2,$$
$$\Delta_x=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ 1 & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$
$$\Delta_y=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & 1 & 1\\ k & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$
$$\Delta_z=\left|\left(\begin{array}{cc}1 & 1 & 1\\1 & k & 1\\ k & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$ which gives the answer:
i) $k\neq-2$ and $k\neq1$;
ii) $k=1$;
iii) $k=-2$.
