MHB Is a Matrix Invertible and Have Integer Entries If Its Determinant is 1 or -1?

[Dethrone]

Let $A$ be a matrix each of whose entries are integers. Prove that $A$ is invertible and $A^{-1}$ having integer entries if and only if $\text{det}A=1$ or $-1$.

What I have so far is this:
Suppose $A$ is invertible, then $AA^{-1}=I$ and $\text{det}A\cdot \text{det}A^{-1}=1$.
I am not sure how to proceed.

 

[Dethrone]

I think I figured it out, but I'm too lazy to type out the formal proof with latex. The forward direction can be proven directly by seeing that if $A$ has integer entries, then the adjugate/adjoint will have integer entries. Furthermore, since $A^{-1}$ has integer entries, $\text{det}A$ must either be 1 or -1. Approaching this by contrapositive, we can see that if $\text{det}A$ is not 1 or -1, then $A^{-1}$ has fractional entries. A contradiction.
The backward direction is just reverse logic. Is that correct?

 

[Jameson]

I agree with the forward direction. $$A^{-1}=\frac{1}{\text{det }(A)} \text{adj }(A)$$ and $\text{adj }(A)$ will contain all integers for this particular $A$ (might need to justify this depending on how in depth your prof likes you to go), so clearly $$\frac{1}{\text{det }(A)}$$ must be an integer in order for the $A^{-1}$ to contain only integers, and the result follows.

For the backward direction, you are suggesting starting with $\displaystyle {\text{det }(A)}= \pm 1 \implies \frac{1}{\text{det }(A)} \text{adj }(A) $ contains all integer entries $\implies A^{-1}$ contains all integer entries. Right?

I think this proof is solid but saying some words about the adjunct matrix containing all integers would be good because it is a major part of your argument in both directions. :)

 

[Dethrone]

Yep, that is precisely what I wanted to say. On paper, I like to be really rigorous and thorough with my explanation, but having to use \text{} each time was too much for me to handle. :p Thanks Jameson!

 

[I like Serena]

$\DeclareMathOperator{\adj}{adj}$

I agree with the forward direction. $$A^{-1}=\frac{1}{\text{det }(A)} \text{adj }(A)$$ and $\text{adj }(A)$ will contain all integers for this particular $A$ (might need to justify this depending on how in depth your prof likes you to go), so clearly $$\frac{1}{\text{det }(A)}$$ must be an integer in order for the $A^{-1}$ to contain only integers, and the result follows.

Yep, that is precisely what I wanted to say. On paper, I like to be really rigorous and thorough with my explanation, but having to use \text{} each time was too much for me to handle. :p Thanks Jameson!

Hey Rido! ;)

I don't think the proof is valid yet. What if $\adj A$ contains only even numbers and $\det A = 2$?

Btw, \det already is an operator name, and $\adj$ can be written as \operatorname{adj}, which takes care of proper spacing.
Alternatively, if you type \DeclareMathOperator{\adj}{adj} once (as I did in this post), you can use \adj in the remainder of the post. (Nerd)
(It might be a nice improvement if \adj was predefined in our forum.)Here's a different way to proceed, which does not require \text or \operatorname at all.
Suppose $\det A \ne \pm 1$.
Since $\det A \cdot \det A^{-1} = \det I = 1$, one of them must have a magnitude greater than 1, while the other one must be between 0 and 1.
Is that possible for integer matrices?

 

[Jameson]

Good catch, ILS. You're 100% correct. If all the terms contain a common factor then $$\frac{1}{\text{det }(A)}$$ doesn't need to be 1.

Much better idea for the proof as well. :)

 

[Dethrone]

Hi ILS! (Cool)

That makes sense! Indeed, since they have integer elements, the determinant must be larger than 1. It could be 0, but then that would violate the fact that $A$ is invertible, so that could not happen.
Seems like $\adj$ is predefined now...:D

Spoiler

It would also be nice if $\operatorname{col} A,\operatorname{row}A, \operatorname{span}A, \operatorname{rref}A$, etc could be predefined too...hehe. Not sure if that would take up too much necessary space, however.