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?
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 $k \in (-\infty,\frac{1}{3})\cup (\frac{1}{3},\infty).$