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It is known that this is true for any field but since it is specifically mentioned that field is finite, perhaps there is a different proof in finite case.

Tony Pizza
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Jacob Martina
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    It is a fact that if $AB=I$ then $BA=I$ is true for any field (for two square matrixes). What it means is that if A has a right inverse then A has a left inverse and it happened to be the same matrix. Read here for more details https://en.wikipedia.org/wiki/Invertible_matrix#Properties – Nadav Kalma Oct 19 '23 at 05:47
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    The proof you link to works when entries in any field. so one does not need to use that the field is finite. Perhaps there is a different proof which relies on there being finitely many elements in the ground field... I myself don't know of one but someone here may. – coffeemath Oct 19 '23 at 05:49
  • Surely a duplicate – Peter Oct 19 '23 at 05:54
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    See this one, not only this is true for matrices over any commutative ring, this actually holds for even "linear transformations" for any (not necessarily free) finitely generated modules over a commutative ring. – Just a user Oct 19 '23 at 06:07
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    Does this answer your question? If $AB = I$ then $BA = I$ – wnoise Oct 19 '23 at 16:42
  • Hey @wnoise, thanks for the recommendation. I came to one of the solutions in that link. But I had an extra assumption which says "elements of A and B are a member of finite field" but i couldn't use that in my solution. that is what I've asked about. – Jacob Martina Oct 19 '23 at 17:03

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Since there are only finitely many $n×n$ matrices over $F$, for some $i>j$ we will have $A^i = A^j$. Since we are given $AB = I$ we can multiply above equation by $B$ from right repeatedly to obtain $A^{i-j} = I$. And one more right multiplication will give you $B = A^{m - 1}$ where $m = i - j > 0$. From this we get $BA = A^{m-1}A = A^m = I$.

Tony Pizza
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    Now this has become a "Here is the answer, guess the question" game – Stef Oct 19 '23 at 18:21
  • @Stef lol. The question was $A, B$ are matrices over a finite field $F$ such that $AB = I$, prove that $BA = I$. This is actually true for any field and it should have many answers right here on this website. But the finite field condition in the question made it look like possibly there is simpler proof in finite case and indeed there is. – Tony Pizza Oct 20 '23 at 05:18
  • It was an interesting question that most people have probably not seen before. The op shouldn't have edited it out. I wonder what Bolivia contest he is talking about and why does it make him postpone this question. – Tony Pizza Oct 20 '23 at 05:20
  • No , it has become a "I answer every off-topic question to collect rep-points" - game. This question violates the rules in several ways : no effort , question in the title , not in the body and no actual context. – Peter Oct 22 '23 at 16:22
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    @Peter did you even care to read the original question? At least care to read the comments. Read coffeemath's comment to the question. Read at least anything before commenting. For the question it's not my fault op removed it. You are acting like a kid although you definitely aren't. – Tony Pizza Oct 23 '23 at 09:01
  • Then , the proper action is to roll it back. – Peter Oct 23 '23 at 13:20
  • @Peter why? Is it hurting anyone by being here? One can know what was the question by reading the comments so it shouldn't cause any major confusion and potentially hurt someone or something like that. – Tony Pizza Oct 23 '23 at 14:14
  • Moreover I have now edited the question to explain the context so now I don't see any possible reason for you to have problem with my answer. – Tony Pizza Oct 23 '23 at 14:17