The proof of the method outlined by koala is quite simple.
The method claims that if a sequence L of row operations turns a square matrix A into the identity then applying that same sequence L to the identity yields A^{1}.
To prove this, we are first going to consider the matrices L_{ij}(λ), which are such that:
l_{ij} = λ
if p=q then l_{pq}=1
else l_{pq}=0
so for example, if we are dealing with 3 x 3 matrices, then
/1 0 4\
L_{13}(4)=0 1 0 
\0 0 1/
Let us now consider the
product B=L
_{ij}(λ)A.
We know our
summation convention:
b
_{pq}=l
_{pk}a
_{kq}
 if we have i=j then :

if p = i then b_{iq}=l_{i k}a_{kq}
l_{i k} = 0 if i ≠ k, therefore b_{iq }= l_{ii}a_{iq}=λa_{iq} 
if p ≠ i then the above remains true, except that l_{pp} = 1, therefore b_{pq } = a_{pq }
From this it is clear that multiplying A by L_{ii}(λ) multiplies the ith row of A by λ and leaving the other rows untouched.

if i ≠ j then we can see that if p = i then
b_{iq}=l_{ik}a_{kq}
=l_{ii}a_{iq} + l_{ij}a_{jq}=a_{iq} + λa_{jq}
multiplying A by L_{ij}(λ) is adding λ times the jth row of A to the ith row, leaving the rest untouched.
Thus we can perform any of the row operations ""multiply a row by λ" and "add λ times a row to another row" by multiplying A by some matrix L
_{ij}(λ). Note also that for all the row operations used in this method the L
_{ij}(λ) are
triangular with non zero terms on the diagonal : they are
invertible.
We need one more type of matrix : one that will allow us to swap the ith and jth row. Fortunately, this is not too difficult. we simply have S_{ij}(with i ≠ j) which is such that :
s_{ij} = 1
s_{ji} = 1
s_{ii} = 0
s_{jj} = 0
and all the other terms on the diagonal are 1. You may wish to check for yourself that this matrix does what I claim, and that it is invertible.
We can now perform all the row operations needed by multiplying by an invertible matrix. So what our method is actually doing is finding a series of m invertible matrices such that L_{m}...L_{2}L_{1}A= I
L_{m}...L_{2}L_{1} represents our sequence of row operations. Multiplying on the right side by A^{1} yields L_{m}...L_{2}L_{1}I= A^{1}, which is what we wanted : performing our sequence of row operations on the identity gives us A^{1}.
You may also be interested in knowing what happens if this method is applied to a singular matrix. What will happen is that eventually you will have a row with nothing but zeros, it will not be possible to transform the matrix into the identity.