# Determinant/Diagonal block matrix/Left lower block 0/Exercise

Let ${\displaystyle {}M}$ be the following square matrix
${\displaystyle {}M={\begin{pmatrix}A&B\\0&D\end{pmatrix}}\,,}$
where ${\displaystyle {}A}$ and ${\displaystyle {}D}$ are square matrices. Prove that
${\displaystyle {}\det M=\det A\cdot \det D}$.