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Let $A$ be a $n \times n$ matrix with integer entries. Prove that all entries of $A^{-1}$ are integer if and only if $\det (A) = \pm 1$.

I know that $$A^{-1}= \frac{\operatorname{adj}(A)}{\det (A)}$$ but I have no idea where to go from there for the forward direction. Any help here would be greatly appreciated.

For the backwards direction, I think I am okay. I plugged the $1$ and the $-1$ options into $\frac{\operatorname{adj}(A)}{\det (A)}$. So, I know that $A^{-1} = \pm \operatorname{adj}(A)$. Then I said that because the ${\rm adj}(A)$ is simply the matrix of co-factors of $A$, and because $\operatorname{adj}(A)$ has all integer entries, then $\operatorname{adj}(A)$ will have all integer entries, which will mean that $A^{-1}$ will have all integer entries. Is that okay? Or am I missing something else?

Rodrigo de Azevedo
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Theresa
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2 Answers2

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The direction you've done is correct to a t and is exactly how I'd do it. As for the other direction, you might be tempted to try to reverse your argument but that definitely isn't going to go your way. Here's a hint: suppose that $A^{-1}$ has integer entries, then we know that

$$\det(A^{-1}A) = \det(I) = 1$$

but we also have that $\det(A^{-1}A) = \det(A^{-1})\det(A)$. Can you take it from here?

Cameron L. Williams
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  • Okay, so that would give me $1 = det(A^{-1})det(A)$ so $det(A) = 1/det(A^{-1})$. but where would I go from there? I just don't think I am seeing this one. Sorry about the notation too haven't figured that out yet.. – Theresa Apr 21 '15 at 00:07
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    @Theresa You know that $A$ and $A^{-1}$ are matrices with integer entries; what does that say about their determinants? – Steven Stadnicki Apr 21 '15 at 00:11
  • @StevenStadnicki It would mean that their determinants are also integers..But I don't see how to ensure $1/det(A^{-1})$ is in fact +/- 1. – Theresa Apr 21 '15 at 00:25
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    @Theresa you've got the right idea. Both $\det(A)$ and $\det(A^{-1})$ are integers as you note and their product is $1$. What integers can you multiply to get $1$? – Cameron L. Williams Apr 21 '15 at 00:26
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    Okay, I see it. I am surprised I didn't see it before. Thank you!!! – Theresa Apr 21 '15 at 00:28
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    @Theresa It's easy to overlook and it took me a few minutes of thinking to put it together the right way. Glad you got it! :) For a first question on the site, you did very well by adhering to our guidelines (not many new users do!). Please do not hesitate to ask questions in the future or even answer those you feel that you can answer. – Cameron L. Williams Apr 21 '15 at 00:28
  • @CameronWilliams I'm working on a problem that might have something to do with this, except it's in the area of free abelian groups. Would you be willing to take a look at it? I have a 50 point bounty on it: http://math.stackexchange.com/questions/2120233/x-is-a-basis-for-free-abelian-group-a-n-if-and-only-if-det-m-pm-1 –  Feb 02 '17 at 19:05
  • @JessyunBourne Oh goodness. Free groups aren't my wheelhouse at all, but I'll check it out. No guarantees at all though. – Cameron L. Williams Feb 03 '17 at 02:09
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You did one direction well. Here the other direction:

If $A$ only contains integers then $\det(A)$ is an integer (computing a determinant only uses $+$ and $\cdot$). Assume $A^{-1}$ is also an integer matrix, then $\det(A^{-1})$ is also integer. But

$$\det(A^{-1})=\frac 1{\det(A)}$$

can be an integer only if $\det(A)\in\{-1,1\}$ because $\pm 1$ are the only integers with integer reciprocals.

M. Winter
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