Finding the position of two lines for each value of k I have the two lines r and s
$$ r: \begin{cases} x+y=1 \\x+z=1 \end{cases} $$ $$ s: \begin{cases} x-ky=k \\z-x=k \end{cases}$$ 
First, I create one single system with all equations
$$ \begin{cases} x+y=1 \\x+z=1\\x-ky=k\\x-z=k \end{cases} $$
which I rewrite as
$$\begin{matrix} 1 & 1 & 0 & 1 \\1&0&1&1\\1&-k&0&k\\1&0&-1&k\end{matrix}$$
I then transform it in row echelon form, getting
$$ \begin{matrix} 1&1&0&1\\0&-1&1&0\\0&0&-k-1&k-1\\0&0&0& \frac{-(k-1)^2}{-k-1} \end{matrix} $$
I find the values of k by getting the determinant of the matrix
$$ -(k-1)^2 = 0 \rightarrow k = 1  $$
So if I write the ref matrix with 1 instead of k I get
$$\begin{matrix} 1&1&0&1\\0&-1&1&0\\0&0&-2&0\\0&0&0&0 \end{matrix} $$
The matrix has rank 3, which is not its maximum rank, meaning the system has infinite solutions.
If the system has infinite solutions, does that means that the lines are always parallel when k=1? And since the rank of the matrix is 4 for each value of k other than 1, does that mean that the two lines always intersect when k is not 1?
 A: You have to solve the system
$$r=s$$
$$1-r=-1+\frac{s}{k}$$
$$1-r=k+s$$ where $r,s$ are real numbers.
From the first equation we get $$r=s$$, so we get
$$s=\frac{1-k}{2}$$ and the last equation gives $$2=s\left(\frac{1}{k}+1\right)$$
We have $$s=\frac{1-k}{2}$$ and $$s=\frac{2k}{k+1}$$ so we get one solution if $$\frac{2k}{k+1}=\frac{1-k}{2}$$ and you have to solve the equation $$k^2+4k-1=0$$
The case $k=0$ gives $$x=0,y\in\mathbb{R},z=0$$
A: You made a sign error when you combined the two systems of equations: the second equation for $s$ is $z-x=k$, but the last equation in your combined system is $x-z=k$, which is not the same thing. If you reorder the variables, it should be either $-x+z=k$ or $x-z=-k$. This error cascades through the further calculations.  
With the correct equation, the last row of the reduced matrix is $$\begin{bmatrix}0&0&0&{k^2+4k-1\over k+1}\end{bmatrix}.$$ This system is therefore consistent, and so has an infinite number of solutions as you stated, when $k^2+4k-1=0$. (As it turns out, the determinant of the reduced matrix is also equal to $k^2+4k-1$, but there’s not really any need to compute it.) The lines are coincident, not parallel, for those values of $k$. If they were parallel, there would be no solutions.  
You also need to examine the case $k=-1$ separately since in that case the row-reduction you performed is invalid because of a division by zero. After substituting this value of $k$ into the original equations, one can see at a glance that the system is inconsistent: for $r$ you have $x+y=1$, but for $s$ you have $x+y=-1$. The planes defined by these two equations are parallel, so there are no points in common between the two lines at all.  
Your conclusion that the system is consistent and therefore has a single solution for other values of $k$ than the three above is incorrect, however. The rank of the augmented matrix is indeed $4$, but the rank of the coefficient matrix cannot be greater than $3$, so the system has no solutions for these values of $k$. To put it another way, the last row of the reduced matrix above represents the equation $$0={k^2+4k-1\over k+1}$$ (and a similar equation for your erroneous computation), which has no solutions unless the right-hand side vanishes.  
Finally, having no intersection in 3-D doesn’t mean that the lines are parallel as it does in two dimensions. The lines might be skew: they simply pass by each other in different directions.
