Product of a matrix and its adjugate

Statement

Let $K$ be a commutative ring, let $n \ge 2$, and let $A$ be an $n \times n$ matrix over $K$. Then $$A \operatorname{adj}(A) = \operatorname{adj}(A) A = \det(A)\, I_n,$$ where $\operatorname{adj}(A)$ is the adjugate of $A$ and $\det$ is the determinant.

Remarks

The diagonal entries of $A \operatorname{adj}(A)$ are the Laplace expansions of $\det A$ along the rows. The entries off the diagonal are expansions of matrices with two equal rows, so they are zero. The identity holds for all square matrices, also for singular ones.

Explanations

L1 human

Multiplying a matrix by its adjugate gives the determinant times the identity matrix. So, if the determinant is invertible, then dividing the adjugate by it gives the inverse. For a singular matrix, the product is the zero matrix.