Determinant of a block triangular matrix

Statement

Let $K$ be a commutative ring, let $A$ be an $m \times m$ matrix, $B$ an $m \times n$ matrix and $D$ an $n \times n$ matrix over $K$. Then the block matrix satisfies $$\det \begin{pmatrix} A & B \\ 0 & D \end{pmatrix} = \det A \cdot \det D,$$ where $0$ is the $n \times m$ zero matrix and $\det$ is the determinant.

Remarks

The same holds for the lower block triangular matrix $\begin{pmatrix} A & 0 \\ C & D \end{pmatrix}$. It generalizes the rule for triangular matrices. For a general block matrix, $\det \begin{pmatrix} A & B \\ C & D \end{pmatrix} \ne \det A \det D - \det B \det C$ in general.

Explanations

L1 human

When a matrix has a block of zeros in its lower left corner, its determinant splits. It is the product of the determinants of the two diagonal blocks. The block $B$ does not matter, just like the entries above the diagonal of a triangular matrix.