Find all the values of a for a system that has a) no solution b) 1 solution c) infinitely many solutions Solving for a system for a with a matrix
x + ay - z = 1
-2x - ay + 3z = -4
-x -ay + ay = a + 1

This is the solution I found:
1 a -1   1
0 a  1  -2
0 0  a-1 a+2

And reduced it even further

   1  0 -2    3
   0  1  1/a  -2/a
   0  0  a-1   a+2


1. Is it possible to have inifinite solutions? Would no solution be when  a = 1 or a = -2?
2. What is also confusing is in my first solution, the first row works with a =1 but not a = -2? Does this mean that a = -2 is not a solution.
I apologize about my formating if someone can let me know how to proper format my matrixes that would be great. 

You can only perform your final reduction if $a\ne0$; so you need to
split off $a=0$ as a separate case and investigate it individually.
You also point out the cases $a=1$ and $a=-2$. Why don't you investigate
these too? For instance when $a=1$ you get the final row $0\ 0\ 0\ 3$
which is the equation $0x+0y+0z=3$: impossible. But what happens with $a=-2$?
