Determine when the system has a) no solution, b) 1 solution and c) infinitely many solutions This question is not for an assignment, it was on the midterm and I am interested in figuring out how to solve it before the final exam. 
cheers,
Determine when the system has a) no solution, b) 1 solution and c) infinitely many solutions
Given the system
\begin{align*}
x + y + 7z & = -7\\
2x + 3y +17z & = -16\\
x +2y +(a^2+1)z & = 3a
\end{align*}
Attempted Solution
using matrix form $$\pmatrix{1&1&7&-7\cr2&3&17&-16\cr1&2&a^2+1&3a\cr}$$ using gaussian elimination to isolate eqn 3 for variable z (sorry about the missing a^2+1 up there html doesn't like me)
line 1 - line 3
$$\pmatrix{1&1&7&-7\cr2&3&17&-16\cr 0&1&7-(a^2+1)&-7-3a\cr}$$
2*line 1 - line 2
$$\pmatrix{1&1&7&-7\cr 0&-1&-3&2 \cr 0&-1&7-(a^2+1)&-7-3a\cr}$$
-1*line2
$$\pmatrix{1 & 1 &7 &-7\\ 0 & 1 & 3 & -2\\ 0 & -1 & 7 - (a^2 + 1) & -7 - 3a}$$
-1*line 2 + line 1
$$\pmatrix{1 & 0 & 4 & -5\\ 0 & 1 & 3 & -2\\ 0 & -1 & 7 - (a^2 + 1) & -7 - 3a}$$
line 2 + line 3
$$\pmatrix{1 & 0 & 4 & -5\\ 0 & 1 & 3 & -2\\ 0 & 0 & 10 - (a^2 + 1) & -9 - 3a}$$
so we get equation 3
$10-(a^2+1)z=-9-3a$
rearranging,
$$(a^2+1)z=-19-3a \rightarrow$$
$$z=19/a^2+1 - 3a/a^2+1 \rightarrow $$
$$z=(19/a^2+1) - (-3/a+1)$$
$$z=19/a^2 + 1 - (-3/a) +1$$
$$z=19/a^2 + 2 + 3/a$$
multiply all terms by $a$
$$za=19/a + 2a + 3$$
$$za=(39/19)a + 3$$
$$za=3a+3$$
$$z=3+3/a$$
wut?
I am unsure how to properly answer the question, please help stack exchange!
My algebra skills are a little lacking, please point out helpfully where my thinking is wrong.
I can solve for z but what does that prove
 A: You have arrived to this (which is followed by some not very reader friendly part, which I am ignoring EDIT: In the meantime, N. F. Taussig fixed it):
$$\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&-1&7-(a^2+1)&-7-3a
\end{array}\right)=
\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&-1&6-a^2&-7-3a
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&0&9-a^2&-9-3a
\end{array}\right)=
\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&0&(3-a)(3+a)&-3(3+a)
\end{array}\right)
$$
For $a=-3$ you have
$$\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&0&0&0
\end{array}\right)$$
This system has infinitely many solutions.
If $a\ne-3$, you can divide by $a+3$ to get
$$
\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&0&3-a&-3
\end{array}\right)
$$
If $a=3$, then it has no solution. (The last equation is $0x+0y+0z=3$.)
For $a\ne3$ you can see that there is only one solution. (Maybe it would be better if I wrote for $a\ne\pm3$; just to remind that now we only consider the case when $a\ne-3$ and we got an additional condition $a\ne3$ to get unique solution.)

As a sanity check, you can plug $a=3$ in the original system. You can notice that if you subtract the first equation from the second, you get
$$x+2y+10z=-9.$$
For $a=3$ the third equation is
$$x+2y+10z=9.$$
These two equations are clearly incompatible.
Also for $a=-3$ you can see that the third equation becomes
$$x+2y+10z=-9.$$
So in this case the third equation is direct consequence of the first two equations.

I will also point out that if I was doing the Gaussian (or Gauss-Jordan) elimination, I would probably start by subtracting the first row from the second, since then I have two very similar rows, which makes things easier).
$$\left(\begin{array}{ccc|c}
  1 & 1 & 7 &-7 \\
  2 & 3 &17 &-16\\
  1 & 2 &a^2+1& 3a
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 7 &-7 \\
  1 & 2 &10 &-9\\
  1 & 2 &a^2+1& 3a
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 7 &-7 \\
  1 & 2 &10 &-9\\
  0 & 0 &a^2-9& 3a+9
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 7 &-7 \\
  1 & 2 &10 &-9\\
  0 & 0 &(a-3)(a+3)& 3(a+3)
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 7 &-7 \\
  0 & 1 & 3 &-2\\
  0 & 0 &(a-3)(a+3)& 3(a+3)
\end{array}\right)$$
