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Mar 9, 2013 at 1:31 comment added Sungjin Kim With help of Benjamin Young's and Brendan McKay's answers, I found that Brendan's guess is true without $n$ on the exponent. Then tried understanding the combinatorial meaning behind it. The formula $AM=I$ is essentially $(1-1)^m=0$ for $m>0$.
Mar 8, 2013 at 10:07 vote accept Yemon Choi
Mar 8, 2013 at 10:07
Mar 8, 2013 at 8:33 history edited Sungjin Kim CC BY-SA 3.0
Added a comment about inverse.
Mar 8, 2013 at 1:58 comment added Yemon Choi Apologies for vagueness, your comment is a fair one. I have updated the question with an attempt to clarify.
Mar 8, 2013 at 0:54 comment added Benjamin Steinberg I gave this a +1. It is no less conceptual than computing the inverse by induction and in a sense it is very much like the first answer except without determinants. I liked all the answers.
Mar 7, 2013 at 19:48 comment added Sungjin Kim The word "conceptual" is really vague in your original posting. What I understand is that my solution is involving the "concept" of "elementary row and column operations don't change the rank of matrix". Here is what I see: Kevin's answer is involving "invertible <=> nonzero determinant" Benjamin Steinberg's answer is involving "change of basis" Benjamin Young's answer is involving "the very definition of invertibility". So, I don't see why mine is specifically what you don't want to see.
Mar 7, 2013 at 19:00 comment added Yemon Choi I explicitly said I wanted a more "conceptual" argument. Of course induction is involved somewhere, but what I wanted was some explanation of how this example might fit into a bigger picture
Mar 7, 2013 at 10:07 history edited Sungjin Kim CC BY-SA 3.0
added 12 characters in body
Mar 7, 2013 at 9:53 history answered Sungjin Kim CC BY-SA 3.0