Inverse in terms of the adjugate

Statement

Let $K$ be a commutative ring, let $n \ge 2$, and let $A$ be an $n \times n$ matrix over $K$ such that $\det A$ is a unit of $K$. Then $A$ is invertible, and $$A^{-1} = (\det A)^{-1} \operatorname{adj}(A),$$ where $\operatorname{adj}(A)$ is the adjugate of $A$.

Remarks

The statement follows by dividing $A \operatorname{adj}(A) = \operatorname{adj}(A) A = \det(A) I_n$ by $\det A$. For $n = 2$ and $ad - bc \ne 0$ over a field, $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. For large $n$, row reduction computes the inverse much faster than this formula.

Explanations

L1 human

This formula gives every entry of the inverse as a quotient of two determinants. It is useful for theory and for small matrices. For $2 \times 2$ matrices, it is the formula that most students learn first.