How to find the values of a matrix that make it consistent? So I have this system of equations:
\begin{eqnarray}
2x+ky+z&=&1,\\
x+z&=&0,\\
2x+y-z&=&k.
\end{eqnarray}
If I put this system in the form of an augmented matrix, I get
\begin{equation*}
\begin{bmatrix}
2    &k&    1&    1\\ 
1   & 0&    1&    0  \\
2   & 1&    -1&    k
\end{bmatrix}
\end{equation*}
I then plug this augmented matrix in my calculator to get a row echelon form , which results in :
\begin{equation*}
\begin{bmatrix}
1   & k/2&    1/2&    1/2 \\
0&    1&    -1/k&    1/k \\
0&    0&    1&    -k²+1/3k-1
\end{bmatrix}
\end{equation*}
I want to find the values of $k$ which make the system consistent. 
Thus, I was thinking of taking into consideration the last row, which implies that $-k²+1/3k-1.$
So my question is : taking into consideration the denominator of $-k²+1/3k-1$, is the system inconsistent where $k=1/3$ (where the denominator is equal to $0$). Should I also seek inconsistency where the numerator is equal to zero?
 A: This calculator (https://www.dcode.fr/matrix-row-echelon) gives the following augmented (reduced) matrix (so check whether your reduced matrix is correct):
\begin{equation*}
\begin{bmatrix}
1 &0 &0 &\frac{k^2-1}{3k-1}\\
0 &1 &0 &-\frac{k-3}{3k-1}\\
0 &0 &1 &-\frac{k^2-1}{3k-1}.
\end{bmatrix}
\end{equation*}
Your system is consistent only if the rank of your coefficient matrix is equal to that of the augmented matrix.  As the rank of the coefficient matrix is equal to $3$ (which is easy to see), the rank of the augmented matrix must be equal to $3$ for consistency.  It is easy to see that as soon as the elements of the fourth column are determined, the rank of the augmented matrix is $3$ too.  Since all elements of the fourth column are fractions with denominator 
$$3k-1,$$
they are determined for all value of $k$ such that the denominator is not equal to zero, i.e.:
$$3k - 1 \neq 0,$$ 
or
$$k \neq \frac{1}{3}.$$
Then, the answer is: the system of linear equations is consistent for $k \in (-\infty,\frac{1}{3})\cup (\frac{1}{3},\infty).$
