Take for example the following matrix:
$$
A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}
$$
The elementary matrix equivalent to changing the first row with the second is
$$
E = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}
$$
multiplied from the left. The elementary matrix equivalent to changing the first column with the second is the same matrix $E$ multiplied from the right.
After a quick check, I found that $E = E^{-1}$.
Given that, I concluded that:
$$
B = \begin{pmatrix} 5 & 4 & 6 \\ 2 & 1 & 3 \\ 8 & 7 &9\end{pmatrix} = EAE^{-1}
$$
and therefore $A\sim B$.
Is this comprehensive? Does changing rows and column necessarily make the outcome similar to the original?
Thanks
Best Answer
Yes, this is correct. If you change rows and columns with this same format, the outcome would be similar to the original.
There is one typo in your $B$ though. The last row should be in the order of $\quad 8\quad 7\quad 9$.