Let $A\in \mathbb{R}^{n\times n}$ be invertible such that $A$ and $A^{-1}$ have integer entries. Prove that $\det(A)\in\{-1,1\}$.

Question

My work:

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

I'm stuck here. Can someone help me?

First piece of help: A^-1 gives the wrong output; you want A^{-1}. — vadim123, Jan 28 '18 at 02:32
Second piece of help: https://en.wikipedia.org/wiki/Minor_(linear_algebra)#Inverse_of_a_matrix — vadim123, Jan 28 '18 at 02:33

Tsemo Aristide · Answer 1 · 2018-01-28T02:34:44.887

5

$det(A)$ and $det(A^{-1})$ are integers, $det(AA^{-1})=1=det(A)det(A^{-1})$ and an integer is invertible for the multiplication if and only if it is equal to $1$ or $-1$.

edited Jan 28 '18 at 02:34

answered Jan 28 '18 at 02:33

Tsemo Aristide

89,587

It would be clearer to say the multiplicative inverse of an integer is again and integer iff ... – Ted Shifrin Jan 28 '18 at 02:34
I dont have very clear the part "an integer is invertible for the multiplication if and only if it is equal to $1$ or $−1$" for example, let $a\in \mathbb{Z}$ then $a$ is invertible if $a=1$? – rcoder Jan 28 '18 at 02:41
if a is 1 or -1 – Tsemo Aristide Jan 28 '18 at 02:42

Let $A\in \mathbb{R}^{n\times n}$ be invertible such that $A$ and $A^{-1}$ have integer entries. Prove that $\det(A)\in\{-1,1\}$.

1 Answers1