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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.

  1. $A^{-1} \times \{E_1 \times ... \times E_n\} = A$
  2. $E_{ij} \times A^{-1} \times \{E_1 \times ... \times E_n\}= E_{ij} \times A$
  3. $E_{ij} \times A^{-1} \times \{E_1 \times ... \times E_n\}= B$
  4. $E_{ij} \times A^{-1} \times \{E_1 \times ... \times E_n\} \times \{E_1 \times ... \times E_n\}^{-1}= B \times \{E_1 \times ... \times E_n\}^{-1}$
  5. $E_{ij} \times A^{-1} = B \times \{E_1 \times ... \times E_n\}^{-1}$
  6. $E_{ij} \times A^{-1} = B \times \{E_n^{-1} \times ... \times E_1^{-1}\}$
  7. $F_{ij} = E_{ij}$
  8. $F_{ij} \times A^{-1} = B \times \{E_n^{-1} \times ... \times E_1^{-1}\}$
  9. $F_{ij} \times A^{-1} = B^{-1}$

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.

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  • $\begingroup$ If you are referencing the steps by number could you number them for easy reference? $\endgroup$
    – ruler501
    Mar 21, 2014 at 21:51
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    $\begingroup$ @ruler51 Sure; I had some trouble getting the MathJax to look right when I tried that at first. After I add numbers, I'd appreciate any formatting tips that would make this question look neater. $\endgroup$
    – mttp
    Mar 21, 2014 at 21:55
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    $\begingroup$ @metacompactness Thanks for the formatting assistance $\endgroup$
    – mttp
    Mar 21, 2014 at 21:57

2 Answers 2

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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 $B^{-1}$?

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Here is a brute force method, that's kind of just an exercise in keeping track of indices.

Let $A=(a_{kl})$ and $A^{-1}=(c_{kl})$ so we know that

$$\sum_{m=1}^nc_{km}a_{ml}=\delta_{kl}$$

Now assume $B=(b_{kl})$ and $B^{-1}=(d_{kl})$ are as you've described in relation to $A$. Then, the $kl$-th entry of $B^{-1}B$ is

$$\sum_{m=1}^nd_{km}b_{ml}=d_{ki}b_{il}+d_{kj}b_{jl}+\sum_{m\ne i,j}c_{km}a_{ml}$$

The conditions you've described require that $d_{ki}=c_{kj},$ $b_{il}=a_{jl}$, $d_{kj}=c_{ki}$, and $b_{jl}=a_{il}$. Substituting these values you should find

$$\sum_{m=1}^nd_{km}b_{ml}=\sum_{m=1}^nc_{km}a_{ml}=\delta_{kl}$$

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  • $\begingroup$ Why do we need to add $d_{ki}b_{il}+d_{kj}b_{jl}$ to what I presume (maybe incorrectly?) is the identity matrix in the second equation? $\endgroup$
    – mttp
    Mar 21, 2014 at 22:35
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    $\begingroup$ @mttp: In the sum involving the $d$'s and $b$'s, I've separated out those summands which correspond to indices $i$ and $j$. For the summands remaining, we can change to $c$'s and $a$'s without having to change indices, and for those that were separated, we can change to $c$'s and $a$'s only by swapping indices $i$ and $j$. When we make these switches, the resulting two summands fit nicely back into the sum to give us exactly what we want. $\endgroup$
    – Jared
    Mar 21, 2014 at 22:44

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