find values of k that makes it have solutions, no solutions or infinite amount of solutions Linear Algebra 
This was the original question system presented
$x  −  2y  +  4z  =  6$
$x  +  y  +  z  =  k$
$2x  −  y  +  5z  =  k^2$
I got it reduced down to 
\begin{bmatrix}
1 & 0 &  2&-2k+18 \\ 
 0& 1 &  -1& 6-k \\ 
0 & 0 & 0& 3k^2-k-30
\end{bmatrix}
in matrix form
But I don't know how to continue on to find whether or not it has solutions, no solutions or an infinite amount of solutions
 A: I am going to assume you meant the system resolves to this:
$$
\\
[A|b]=\begin{bmatrix}
1 &0  &2  &-2k+18 \\ 
 0& 1 &-1  & 6-k\\ 
 0&0  &0  &  3k^2-k-30
\end{bmatrix}_{REF}$$
A matrix representing a system of equations makes no sense with $x$ in the matrix.
For unique solutions, $\text{rank}[A|b]=\text{rank}[A]=3$.
The easiest way to find the rank of a matrix is to count the number of non-zero rows when in REF.
We see $[A]$ always has a rank of two, so we never have unique solutions.
For infinite solutions, $\text{rank}[A|b]=\text{rank}[A]<3$.
We can see through that as $\text{rank}[A]=2$, so for $\text{rank}[A|b]=2\Rightarrow 3k^2 -k-30=0$.
For no solutions, $\text{rank}[A|b]>\text{rank}[A]$.
This will occur for any $k$ value where $0\neq3k^2-k-30.$
Hope this made sense.
A: From the reduced matrix form, it can be seen that the system will have solution if,
$$ 3k^2-k-30=0 $$ 
A: Note that the reduced matrix form derived is not quite  correct. Add the first two equations 
$$x  −  2y  +  4z  =  6$$
$$x  +  y  +  z  =  k$$
to get
$$ 2x -y +5z = k+6$$
which is ‘parallel’ to the third equation,
$$2x  −  y  +  5z  =  k^2$$
So, the system either has no solution or infinite number of number of solutions. 
For infinite number of solutions, the right-hand-sides of the two equations above are equal,
$$k+6=k^2$$
So, $k$ takes the $k=3,\>\>\>k=-2$ for the system of equations to have infinite solutions. Otherwise, there are no solutions.
