Here's a method for inverting matrices using row reduction, which I think is what you are after:
Let $A$ be an $n\times n$ matrix. Let $B$ be the $n\times 2n$ matrix obtained by placing the $n\times n$ identity matrix to the right of $A$: $B = \left( A\ |\ I_n\right).$
Now, perform row reduction on this matrix until the left half is the identity matrix (or has a row of zeros). Then the matrix on the right half is $A^{-1}$ (or, if you got a row of zeros in the left half, then $A$ is not invertible).
The reason this works is that performing elementary row operations is equivalent to multiplying on the left by an elementary matrix. By doing the same operations to the identity matrix, you are computing the product of those elementary matrices. If $E_1,\ldots,E_n$ are elementary matrices such that $E_nE_{n-1}\cdots E_1A = I_n,$ then it follows that $E_nE_{n-1}\cdots E_1 = E_{n}E_{n-1}\cdots E_1I_n = A^{-1}$. (Because for $n\times n$ matrices over fields, if $CA=I_n$ then $AC=I_n$). Thus, the computations on the right half of the matrix $B$ give you the inverse of $A$.
For large matrices, this is certainly easier to do by hand than using the adjugate; it is probably faster in computers as well.