 Solution

Use block matrix elimination with the invertible square block $A$:The left multiplier is block triangular with identity diagonal blocks, so its determinant is $1$. Taking the determinant of the block-triangular product provesThe matrix $D-C{A}^{-1}B$ is the Schur complement of $A$. The proof needs $A$ invertible, but does not assume the Schur complement or the full matrix is invertible.