Laplace expansion along a column

Statement

Let $K$ be a commutative ring, let $n \ge 2$, and let $A = (a_{ij})$ be an $n \times n$ matrix over $K$. Let $C_{ij}$ be the cofactors of $A$. Then, for each column $j$, $$\det A = \sum_{i=1}^{n} a_{ij} C_{ij} = \sum_{i=1}^{n} (-1)^{i+j} a_{ij} M_{ij}.$$

Remarks

It follows from the expansion along a row, applied to $A^T$, because $\det A^T = \det A$. For example, expansion along column $1$ of $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 6 & 7 \end{pmatrix}$ gives $1 \cdot (28 - 30) = -2$.

Explanations

L1 human

The same expansion works down a column instead of across a row. We choose the column with the most zeros, to save work. Each zero entry removes one smaller determinant from the sum.