---
tag: 0783S9D0ZH
kind: statement
role: proposition
name: Determinant of the inverse
msc: [15A15]
uses: [0C3CFHDN6R, 0TWTS88TWN, 04CPJ6AF00]
sources:
  - url: https://en.wikipedia.org/wiki/Invertible_matrix
    locator: Other properties
---

## Statement

Let $K$ be a commutative ring and let $A$ be an [[invertible|0EB9N0S33C]] $n \times n$ matrix over $K$.
Then $\det A$ is a unit of $K$, and
$$\det(A^{-1}) = (\det A)^{-1},$$
where $\det$ is the [[determinant|00KDZRW4C0]].

## Remarks

The proof applies the [[product rule|04CPJ6AF00]] to $A A^{-1} = I_n$: $\det A \cdot \det(A^{-1}) = 1$.
Over a field, this says $\det(A^{-1}) = 1 / \det A$.

## Explanations

### L1 human

The inverse matrix undoes the matrix, so it must undo its scaling of volume.
If $A$ scales volume by $3$, then $A^{-1}$ scales volume by $1/3$.
So the determinant of the inverse is the inverse of the determinant.