Let A = \(\begin{bmatrix} 2 & 0 & -1 \\[0.3em] 5 & 1 & 0 \\[0.3em] 0 & 1 & 3 \end{bmatrix}\)
To apply elementary row transformations we write:
A = IA where I is the identity matrix
We proceed with operations in such a way that LHS becomes I and the transformations in I give us a new matrix such that
I = XA
And this X is called inverse of A = A-1
Note: Never apply row and column transformations simultaneously over a matrix.
So we have:
Applying R2→ R2 - 5R3