Is this the right solution for an unsolvable system? I have to find the values of $k$ such that this system $\left[
    \begin{array}{ccc|cc}
      1&0&-3&-3\\
      2&k&-1&-2\\
1&2&k&1
    \end{array}
\right]$ is:
a) unsolvable  b) solvable with infinite solutions c) one solution.
After some steps we can see in this matrix $\left[
    \begin{array}{ccc|cc}
      1&0&-3&-3\\
      0&k&5&4\\
0&2&k+3&4
    \end{array}
\right]$ for $k=2$ the second and the thrid row are the same, so if the substract them on row becomes all zeros. (And the system is unsolvable)
The next step: $\left[
    \begin{array}{ccc|cc}
      1&0&-3&-3\\
      0&k&5&4\\
0&0&k^2-7&4k-8
    \end{array}
\right]$ now im not that sure but I think, the solutions to $k^2-7=0$ should be the values of $k$ such that there are infinite solutions.
Then every possbile value k besides 2 and the solutions for $k^2-7=0$ would return a solvable system with one solution?  

from the first equation we get
$$x_1=-3+3x_3$$
plugging this in the second equation and solving for $x_3$ we have
$$x_3=\frac{4}{5}-\frac{k}{5}x_2$$
and for the third equation we get
$$x_2\left(2-\frac{k}{3}(3+k)\right)=4-\frac{4}{5}(3+k)$$
can you finish?
from the last equation we get
$$x_2(6-3k-k^2)=\frac {3}{5}(8-4k)$$
