Find for which values of $k$ the system has none, one or infinitely many solutions Given the following linear system
$$\left[\begin{array}{cccc|c} k &amp; 0 &amp; 1 &amp; 2 &amp; 0\\ -1 &amp; 1 &amp; k &amp; 1 &amp; 0\\ 1 &amp; k &amp; 1 &amp; 3 &amp; 0\end{array}\right]$$
find for which values of $k$ the system has none, one or infinitely many solutions.

I got to
$$\left[\begin{array}{cccc|c}1&amp;k&amp;1&amp;3&amp;0\\0&amp;1+k&amp;1+k&amp;4&amp;0\\0&amp;-k^2&amp;1-k&amp;2-3k&amp;0\end{array}\right]$$
but now I'm stuck. How do I eliminate $-k^2$ using $1+k$?
This is not the first time I got a question like this, how do I solve it?
You’ll have to do a case analysis to proceed with the row-reduction. Dividing the second row by $1+k$ is legal as long as $k\ne1$, which you’ll have to deal with separately. You could instead proceed by swapping the last two rows and dividing the second row by $-k^2$, but that gives you a different special case, $k=0$ to deal with.  

However, the highlighted question can be answered without doing any calculations whatsoever. The system is homogeneous, so it has at least one solution: the zero vector. Moreover, the system is underdetermined, so it has an infinite number of solutions for all $k$.  

Now, if you were asked for the dimension of the solution space for different values of $k$, that would be a different matter. You’d have to do some calculations to find values of $k$ for which the null space of the coefficient matrix is more than one-dimensional. As an alternative to row-reduction, you could examine the $3\times3$ minors of the coefficient matrix. It turns out that there’s no value of $k$ for which they all vanish, so the matrix has full rank for all $k$.
To perform the second step in row reduction, you need to find the first non-vanishing element in the second row.  That would be $k+1$ iff $k\neq-1$.  In that case multiply the second row by $k^2$ and the third one with $k+1$ and add both.

Now if $k=-1$ you get
$$\left[\begin{array}{cccc|c}1&amp;-1&amp;1&amp;3&amp;0\\0&amp;0&amp;0&amp;4&amp;0\\0&amp;-1&amp;2&amp;5&amp;0\end{array}\right],$$
which is easily solved.