The group of invertible matrices with entries in Z

Question

Let $G=GL_{n}(\mathbb{Z})$, the group of invertible matrices with entries in $\mathbb{Z}$. Then show that

$G=\{A\in M_n(\mathbb{Z}) : det(A)=1 \,\,\,or -1\}$

Can you help me please?

Hints: (1) $\det : M_n(\mathbb{Z})\to \mathbb{Z}$ is multiplicative and (2) the adjugate of a matrix with integer entries also has integer entries and the inverse of a matrix is the adjugate scaled by the reciprocal of the determinant. — Amitesh Datta, Jan 18 '14 at 06:04

André Nicolas · Accepted Answer · 2014-01-18T06:08:07.437

4

Hint: The determinant of a product is the product of the determinants. And the determinant of an identity matrix is $1$.

For the other direction, one way is to use the formula for the inverse in terms of determinants.

edited Jan 18 '14 at 06:08

answered Jan 18 '14 at 06:02

André Nicolas

2

Adding to your hint: Product of two integers is 1, then they are both +1 or both -1. – user44197 Jan 18 '14 at 06:04

1 Answers1