Determinant of a scalar multiple

Statement

Let $K$ be a commutative ring, let $A$ be an $n \times n$ matrix over $K$, and let $c \in K$. Then $$\det(cA) = c^n \det A,$$ where $cA$ is the scalar multiple and $\det$ is the determinant.

Remarks

Each of the $n$ rows is multiplied by $c$, and each row contributes one factor $c$. For example, $\det(-A) = (-1)^n \det A$. So an $n \times n$ real skew-symmetric matrix with $n$ odd has determinant $0$.

Explanations

L1 human

Scaling a matrix by $c$ stretches every edge of a box by $c$. In $n$ dimensions, this multiplies the volume by $c^n$. So doubling a $3 \times 3$ matrix multiplies its determinant by $8$.