Determinant is multilinear in the rows

Statement

Let $K$ be a commutative ring, let $n \ge 1$, and let $1 \le i \le n$. Let $A$, $B$, $C$ be $n \times n$ matrices over $K$ that agree outside row $i$. Let $a$, $b$, $c$ be row $i$ of $A$, $B$, $C$, and assume $c = sa + tb$ with $s, t \in K$. Then $$\det C = s \det A + t \det B,$$ where $\det$ is the determinant.

Remarks

This is property 1 of the definition of the determinant. The determinant is not additive in the whole matrix: in general $\det(A + B) \ne \det A + \det B$. By $\det A^T = \det A$, the determinant is also multilinear in the columns.

Explanations

L1 human

If we fix all rows but one, then the determinant is a linear function of the remaining row. So a sum in one row splits the determinant into a sum of two determinants. A factor in one row comes out of the determinant.