Solving Linear Systems with Arbitrary Constants I've run into somewhat of a problem during my Linear Algebra homework and I can't make heads or tails of it for some odd reason. I'm hopeful that one of you could help me out. It's worded as so :
Consider the System of Linear Equations


|x + y - z = 2|
  |x + 2y + z = 3|
  |x + y + (k^2 - 5)z =  k|

where k is an abitrary constant. For which value(s) of k does this system have 
a unique Solution, infinite many solutions and no solutions

I've already looked through the whole chapter, and did a lot of searching but I can't seem to figure out how to prove it. I can tell that if K = 2 then the first and third equations will be the same, but I don't know how I could word that, and I dont know if the second one has to be the same in order to qualify as infinitely many solutions. I feel like I'm missing something. Any help would be appreciated!
 A: Indeed, if $k = 2$, the first and third equations will be the same, so the system of equations will have infinitely many solutions.


*

*To exhaust all such values that lead to infinite solutions, I suggest
you construct the augmented coefficient matrix for you system of
equations, $$\begin{pmatrix} 1 & 1 & -1 &|& 2 \\ 1 & 2 & 1 &|&3\\ 1 & 1 & k^2 - 5 &|& k\end{pmatrix}$$and then put it into row-echelon form (e.g. add $-R_1$ to $R_2$ and to $R_3$). That will help simplify matters.
If there are any values of $k$ at which one or more rows become all
zeros, then there will be infinitely many solutions for each such
values $k$.

*Again, looking at your the augmented coefficient matrix, after row reducing it: If there are values of $k$ at which any row has all zeros except for a
non-zero entry in the right-most column (the augmented portion), then
the system of equations is inconsistent, meaning NO solution exists
at that value of $k$. Note that $k = -2$ is such a value.

*Finally, each and every value of $k$ not leading to infinitely many
solutions and not leading to no solution will give rise to a unique solution.
