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I've been asked the following question: considering that $A$ is an invertible matrix / $A$ and $A^{-1}$ have integer coefficients, why both determinants must be $1$ or $-1$?

We know that, in linear algebra, an $n$-by-$n$ square matrix $A$ is called invertible if there exists an $n$-by-$n$ square matrix $A^{-1}$ such that $AA^{-1}=I$ where $I$ is the identity matrix.

So, if we also consider the following properties: $A$ is invertible $\Leftrightarrow$ $\det(A)\not=0$ and that $\det(I)=1$.

Then, let $A\in\mathbb Z^{n\times n}$ such that $A^{-1}\in\mathbb Z^{n\times n}$ and in consequence, $\det\colon\mathbb Z^{n\times n}\to \mathbb Z$, now we can say:

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

I cannot realize why it could also be $-1$. Any idea or suggestion about how can I prove it?

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  • $\begingroup$ Will an example suffice? $\endgroup$ – Chappers Jun 18 '17 at 23:33
  • $\begingroup$ you are looking for integer roots for the equation $x^2 = 1$. Just knowing that $ \pm I$ has determinant $\pm 1$ suffices. $\endgroup$ – mdave16 Jun 18 '17 at 23:34
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    $\begingroup$ Because $-1 \times -1 = 1$. $\endgroup$ – Jonathan Hebert Jun 18 '17 at 23:34
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    $\begingroup$ So, you learned that $\det(A)\cdot \det(A^{-1})=1$, you learned that $\det(A)$ must be an integer, and so too must $\det(A^{-1})$. You further know that $\det(A^{-1})=\frac{1}{\det(A)}$ from properties of determinants. There are exactly two possibilities then in the integers for factors of $1$, and thus possible values of $\det(A)$ and $\det(A^{-1})$, namely $1$ and $-1$. Alternatively, there are only two possible integers that their reciprocal is also an integer. That both are possible should be obvious from looking at the $1\times 1$ matrices $[1]$ and $[-1]$ respectively. $\endgroup$ – JMoravitz Jun 18 '17 at 23:35
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    $\begingroup$ It's also interesting to prove that the converse is true: if $\det(A) = \pm 1$ and $A$ has integer entries, then $A$ is invertible and $A^{-1}$ also has integer entries. (Hint for proof: adjugate formula for the inverse.) $\endgroup$ – Daniel Schepler Jun 18 '17 at 23:51
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Suppose $A$ is an invertible matrix with integer coefficients such that $A^{-1}$ has integer coefficients. Then, $$AA^{-1}=I\quad\Rightarrow\quad \mathrm{det}(AA^{-1})=\mathrm{det}(I)=1 $$ But since that for all matrices, $ \mathrm{det}(AB)= \mathrm{det}(A) \mathrm{det}(B)$, we have $$ \mathrm{det}(A) \mathrm{det}(A^{-1})=1.$$ You may notice that the formula for the determinant of a matrix only contains addition/substraction and multiplication. This means that if all the entries of a matrix are integers, then the determinant of the matrix is an integer. By hypotesis, we thus have $$\mathrm{det}(A)=m\in\mathbb{Z}^*,\qquad \mathrm{det}(A^{-1})=n\in\mathbb{Z}^*,$$ and $mn=1$. The only solution to this is $m=n=1$ or $m=n=-1$, which is the desired result.

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If matrices $A$ and $A^{-1}$ have only integer coefficients, that means that both of them must have integer-valued determinant.

And by Cauchy–Binet formula we get: $$ det(AA^{-1})=det(A)det(A^{-1})=1=det(I).$$

From here we directly get statement you want to prove.

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$det(A)det(A^{-1})=1$ implies $det(A)$ or $det(A^{-1})$ are divisors of $1$. Hence they can only be 1 or -1.

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