---
tag: 0BC3Q6K92H
kind: statement
role: theorem
name: Matrix determinant lemma
msc: [15A15]
uses: [0C3CFHDN6R, 0TWTS88TWN]
sources:
  - url: https://en.wikipedia.org/wiki/Matrix_determinant_lemma
    locator: Statement
same_as:
  wikidata: Q6787868
  wikipedia: Matrix determinant lemma
---

## Statement

Let $K$ be a commutative ring, let $A$ be an invertible $n \times n$ matrix over $K$, and let $u, v \in K^n$ be column vectors.
Then
$$\det(A + u v^T) = (1 + v^T A^{-1} u) \det A,$$
where $\det$ is the [[determinant|00KDZRW4C0]].

## Remarks

The matrix $u v^T$ has rank at most $1$, so the lemma describes the effect of a rank one update on the determinant.
It follows from [[Sylvester's determinant theorem|0G24VQ1088]] with the matrices $A^{-1} u$ and $v^T$.
A variant without inverses is $\det(A + u v^T) = \det A + v^T \operatorname{adj}(A) u$.

## Explanations

### L1 human

Adding a matrix of rank one, $u v^T$, changes the determinant by a simple factor.
The factor is $1 + v^T A^{-1} u$, a single number.
So we can update a determinant cheaply when we know the inverse.