Why does the augmented matrix method for finding an inverse give different results for different orders of elementary row operations? Why does the augmented matrix method for finding an inverse give different results for different orders of elementary row operations?
Consider the example of elements from the Heisenberg group:
$\begin{bmatrix}1 & a & b\\ 0 & 1 & c\\ 0 & 0 &1 \end{bmatrix}$.
Now, after augmenting this matrix,   
$
  \left[\begin{array}{rrr|rrr}
    1 & a & b & 1 & 0 & 0\\
    0 & 1 & c & 0 & 1 & 0\\ 
    0 & 0 & 1 & 0 & 0 & 1
  \end{array}\right]
$
if you subtract c times the third row from the second,
$
  \left[\begin{array}{rrr|rrr}
    1 & a & b & 1 & 0 & 0\\
    0 & 1 & 0 & 0 & 1 & -c\\ 
    0 & 0 & 1 & 0 & 0 & 1
  \end{array}\right]
$
then b times the third row from the first,
$
  \left[\begin{array}{rrr|rrr}
    1 & a & 0 & 1 & 0 & -b\\
    0 & 1 & 0 & 0 & 1 & -c\\ 
    0 & 0 & 1 & 0 & 0 & 1
  \end{array}\right]
$
and then a times the new second row from the first, 
$
  \left[\begin{array}{rrr|rrr}
    1 & 0 & 0 & 1 & -a & -b\\
    0 & 1 & 0 & 0 & 1 & -c\\ 
    0 & 0 & 1 & 0 & 0 & 1
  \end{array}\right]
$
you then get the following matrix for the inverse:
$$\begin{bmatrix}1 & -a & -b\\ 0 & 1 & -c\\ 0 & 0 &1 \end{bmatrix}.$$
But, this is not the inverse.
I thought order of elementary row operations doesn't matter.

$$\left[ \begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & c  & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$
I believe there is a typo in your working, I believe you mean subtract $c$ times the third row and add it to the second row
$$\left[ \begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & 0  & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$
After which, 
$$\left[ \begin{array}{ccc|ccc} 1 & a & 0 & 1 & 0 & -b \\ 0 & 1 & 0  & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$
and for the last step:
$$\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -a & ac-b \\ 0 & 1 & 0  & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$
Edit: 
Your mistake is at the last step
subtract $a$ times the new second row from the first,
It should involve $(-a)(-c)+(-b)$
