Q:

Proof that matrix $B^{-1}$ = matrix $A^{-1}$ with 2 columns swapped given that B = A with 2 rows swapped.

I'm trying to prove the following. 
Given that $A$ is a nonsingular $n \times n$ matrix, and $B$ is the nonsingular matrix obtained by interchanging rows $i$ and $j$ of $A$, where $i \neq j$, show that $B^{-1}$ can be obtained by interchanging columns $i$ and $j$ of $A^{-1}$. 
Here's my proof so far:
Let $E$ denote a row operation elementary matrix.
Let $F$ denote a column operation elementary matrix.
$E_{ij}$ is the elementary matrix where rows $i$ and $j$ are swapped. 

$A^{-1} \times \{E_1 \times ... \times E_n\} = A$
$E_{ij} \times A^{-1} \times \{E_1 \times ... \times E_n\}= E_{ij} \times A$
$E_{ij} \times A^{-1} \times \{E_1 \times ... \times E_n\}= B$
$E_{ij} \times A^{-1} \times \{E_1 \times ... \times E_n\} \times \{E_1 \times ... \times E_n\}^{-1}= B \times \{E_1 \times ... \times E_n\}^{-1}$
$E_{ij} \times A^{-1} = B \times \{E_1 \times ... \times E_n\}^{-1}$
$E_{ij} \times A^{-1} = B \times \{E_n^{-1} \times ... \times E_1^{-1}\}$
$F_{ij} = E_{ij}$
$F_{ij} \times A^{-1} = B \times \{E_n^{-1} \times ... \times E_1^{-1}\}$
$F_{ij} \times A^{-1} = B^{-1}$

I'm not sure if this is correct. I don't feel comfortable about the way I've transitioned from step 6 to 8 or from step 8 to 9. All hints appreciated.

A:

Hint: There is no difference between "row operation" matrices and "column operation" matrices. Left multiplication by an elementary matrix performs operations on rows, while right multiplication performs operations on columns.
The initial statement of the problem tells you that $E_{ij}A = B$. How can you then write $B^{-1}$? 

