What is wrong in the following calculation for the inverse of a matrix?

Question

$\left[\begin{array}{ccc|ccc} 0 & 3 & 0 & 1 & 0 & 0\\ 4 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 5 & 0 & 0 & 1 \end{array}\right]$

Exchange the first and the second columns:

$\left[\begin{array}{ccc|ccc} 3 & 0 & 0 & 0 & 1 & 0\\ 0 & 4 & 0 & 1 & 0 & 0\\ 0 & 0 & 5 & 0 & 0 & 1 \end{array}\right]$

Divide the rows by $3$, $4$, and $5$ respectively:

$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & \frac{1}{3} & 0\\ 0 & 1 & 0 & \frac{1}{4} & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & \frac{1}{5} \end{array}\right]$

Obviously, the result is not the inverse of the initial matrix. Where is the mistake?

Answer 1

When you use column transformations, this is what you're actually doing:
$AI = A \Rightarrow\\ AIE_1 = AE_1,\ \text{where $E_1$ is an elementary matrix}\\ \vdots\\ AIE_1\ldots E_n = AE_1\ldots E_n = I \Rightarrow\\ IE_1\ldots E_n = A^{-1}$

When you use row transformations, on the other hand, you're left-multiplying by elementary matrices:
$E_1'\ldots E_n'IA = E_1'\ldots E_n'A = I \Rightarrow\\ E_1'\ldots E_n'I = A^{-1}$

You can't combine the two as the same identity matrix cannot be on the left and right of $A$ at the same time, so to speak. And if you use two different identity matrices (as $IAI$, so that the augmented matrix is $[I|A|I]$), then neither one will be transformed into $A^{-1}$.

Answer 2

You shoul work on columns or on rows, never on both.