Question: Here's my system of equations:

$\begin{cases}y + 2kz = 0\\x + 2y + 6z = 2\\kx + 2z = 1\\
\end{cases}$

So I have $\left[\begin{array}{ccc|c}1&2&6&2\\0&1&2k&0\\k&0&2&1\end{array}\right]$

When I row reduce, I get:

$\left[\begin{array}{ccc|c}1&0&6-4k&2\\0&1&2k&0\\0&0&2-6k-4k^2&1-2k\end{array}\right]$

Not really sure where to go from here...

Any suggestions much appreciated!

Thanks,
Mariogs

Answer: 
While DonAntonio's answer is certainly correct, it is likely that your question comes from a class where determinants have not yet been discussed, so you may need a different perspective.

In that case, recall that your system will be inconsistent if, after row reduction, you have a row of the form
$( 0 \ 0 \ 0 \mid 1)$ since this row would correspond to the equation $0x+0y+0z=1$ which clearly has no solutions.

On the other hand, you also don't want a row of of the form $( 0 \ 0 \ 0 \mid 0)$, which would give you a free variable and hence infinitely many solutions.

Thus, **you should find the values of** $k$ **for which** $2-6k-4k^2 = 0$. By our discussion, we can see that as long as you **avoid those values of** $k$, your system will have a solution, and this solution will be unique.

You can find these "bad" values of $k$ by methods from high school algebra, e.g. the quadratic formula.
The coefficients' matrix of your non-homogeneous linear system is:

$$\begin{pmatrix}0&1&2k\\
1&2&6\\
k&0&2\end{pmatrix}$$

and the system has (a unique, by the way) solution iff the above matrix is regular, which happens iff its determinant is **non-zero** ...