This is a very standard method; if you discovered it on your own, congratulations! It works because each of the elementary row operations that you’re performing is equivalent to multiplication by an elementary matrix. To convert $A$ to $I$, you perform some sequence of elementary row operations, which in effect is multiplying $A$ by a sequence of elementary matrices:
$$I=(E_mE_{m-1}\ldots E_2E_1)A\;,$$
say, if it took $m$ row operations. This says that $E_mE_{m-1}\ldots E_2E_1$ is $A^{-1}$. (Well, actually it says that $E_mE_{m-1}\ldots E_2E_1$ is a left inverse for $A$, but there’s a theorem that says that a left inverse of a square matrix is actually the inverse of the matrix.)
You’ve performed exactly the same operations to $I$ on the other side of the augmented matrix, so on that side you end up with
$$E_mE_{m-1}\ldots E_2E_1I=E_mE_{m-1}\ldots E_2E_1\;,$$
which we just saw is $A^{-1}$.
Thus, if you start with $\begin{bmatrix}A\mid I\end{bmatrix}$, you’re guaranteed to end up with $\begin{bmatrix}I\mid A^{-1}\end{bmatrix}$, exactly as you discovered.
In short, row reduced echelon form(RREF) of a matrix $A$ is such that
i) Every leading entry is 1
ii) Any nonzero rows are above zero rows
iii) any leading entry is strictly to the right of any leading entries above that row
iv) any other entry in a column containing a leading entry is 0 except for the leading entry.
So it does not have to be put in augmented matrix $[A|b]$ to get a RRE form. You are comparing RRE form of matrix $A$ and $[A|b]$.
To see why the statement is true, suppose that you put the matrix $[A|b]$ into RRE form, so you have a matrix E. If E contains a leading entry in its last column, in terms of system of equations, what does it say? And what is the condition for E to not have any leading entry in last column?
Note: If RRE form of $[A|b]$ does contain a leading entry, then it is different from that of $A$. Also, note that RRE form of $[A|b]$ is m by n+1 whereas that of $A$ is m by n.
Solve:
$x+y=1$
$x+y=2$
Then we have
$\
A =
\left( {\begin{array}{cc}
1 & 1 \\
1 & 1
\end{array} } \right)
$
$\
b =
\left( {\begin{array}{cc}
1 \\
2
\end{array} } \right)
$
and $Ax=b$
If we turn A into RREF, we get
$\
E =
\left( {\begin{array}{cc}
1 & 1 \\
0 & 0
\end{array} } \right)
$
So A has rank 1
and if we put $[A|b]$ into RRE form, we get
$\
E' =
\left( {\begin{array}{cc}
1 & 1 & 0 \\
0 & 0 & 1
\end{array} } \right)
$
So augmented matrix has rank 2. Observe what last row says in terms of equations.
Best Answer
Simplicity is in the eye of the beholder, but you can calculate the inverse of $A$ using Schur complement. Let $$ A=\pmatrix{X&Y\\ Z&W}, $$ where $X,Y,Z,W$ are square matrices of the same sizes. The matrix $S=X-YW^{-1}Z$ is known as the Schur complement of $W$ in $A$. If both $W$ and $S$ are invertible, then $A$ is invertible too and we have the matrix inverse formula $$ A^{-1}=\pmatrix{S^{-1} & -S^{-1}YW^{-1}\\ -W^{-1}ZS^{-1} & W^{-1} + W^{-1}ZS^{-1}YW^{-1} }. $$ So, to apply this formula to your case, you don't need to calculate any 3x3 or 4x4 adjugate matrices. Instead, you need to compute the inverses and products of a whole bunch of 2x2 matrices.