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Why does the augmented matrix method for finding an inverse give different results for different orders of elementary row operations?

Consider the example of elements from the Heisenberg group: $\begin{bmatrix}1 & a & b\\ 0 & 1 & c\\ 0 & 0 &1 \end{bmatrix}$.

Now, after augmenting this matrix,

$ \left[\begin{array}{rrr|rrr} 1 & a & b & 1 & 0 & 0\\ 0 & 1 & c & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right] $

if you subtract c times the third row from the second,

$ \left[\begin{array}{rrr|rrr} 1 & a & b & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right] $

then b times the third row from the first,

$ \left[\begin{array}{rrr|rrr} 1 & a & 0 & 1 & 0 & -b\\ 0 & 1 & 0 & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right] $

and then a times the new second row from the first,

$ \left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 1 & -a & -b\\ 0 & 1 & 0 & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right] $

you then get the following matrix for the inverse:

$$\begin{bmatrix}1 & -a & -b\\ 0 & 1 & -c\\ 0 & 0 &1 \end{bmatrix}.$$

But, this is not the inverse.

I thought order of elementary row operations doesn't matter.

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  • $\begingroup$ Did you mean "a times the new second row from the first"? If not, then you have not eliminated the $a$ in the original matrix, so you're not done. $\endgroup$ – Ian Sep 24 '17 at 2:13
  • $\begingroup$ @Ian Yes, I did mean first. I'm correcting it now. Thanks! $\endgroup$ – user3146 Sep 24 '17 at 2:43
  • $\begingroup$ Once you do that and also use the new second row, the problem goes away entirely. $\endgroup$ – Ian Sep 24 '17 at 2:46
  • $\begingroup$ I'm not sure what you mean. Are you saying that the resulting matrix that I listed is the inverse of the former? $\endgroup$ – user3146 Sep 24 '17 at 2:55
  • $\begingroup$ No, I'm saying that your third row operation was performed incorrectly. The third row operation should have changed the (1,3) entry again. $\endgroup$ – Ian Sep 24 '17 at 3:21
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$$\left[ \begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & c & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$

I believe there is a typo in your working, I believe you mean subtract $c$ times the third row and add it to the second row

$$\left[ \begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$

After which,

$$\left[ \begin{array}{ccc|ccc} 1 & a & 0 & 1 & 0 & -b \\ 0 & 1 & 0 & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$

and for the last step:

$$\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -a & ac-b \\ 0 & 1 & 0 & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$

Edit:

Your mistake is at the last step

subtract $a$ times the new second row from the first,

It should involve $(-a)(-c)+(-b)$

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  • $\begingroup$ The typo I made was the one Ian suggested. I corrected it. The order of your elementary row operations gives the inverse. I was already aware of this order. This doesn't address my issue. (I suppose it does help to elaborate it though.) Can you please tell me why the 2 different orders (yours and mine) of elementary row operations gives different results for the inverse? $\endgroup$ – user3146 Sep 24 '17 at 2:52
  • $\begingroup$ They don't, unless it's because of an error. $\endgroup$ – G Tony Jacobs Sep 24 '17 at 3:07
  • $\begingroup$ @GTonyJacobs Can you please tell me what error I made? $\endgroup$ – user3146 Sep 24 '17 at 3:12
  • $\begingroup$ When you subtract $a$ times the new second row from the first, you missed that there's a $-c$ in that second row. That should add an $ac$ term to your upper-right entry. $\endgroup$ – G Tony Jacobs Sep 24 '17 at 3:15
  • $\begingroup$ Also, these aren't two different orders of elementary row operations. You both did the same operations, in the same order. $\endgroup$ – G Tony Jacobs Sep 24 '17 at 3:18

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