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.

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 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.