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
Best 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.