# Determinant of block matrix with commuting blocks

I know that given a $$2N \times 2N$$ block matrix with $$N \times N$$ blocks like

$$\mathbf{S} = \begin{pmatrix} A & B\\ C & D \end{pmatrix}$$

we can calculate $$\det(\mathbf{S})=\det(AD-BD^{-1}CD)$$ and so clearly if $$D$$ and $$C$$ commute this reduces to $$\det(AD-BC)$$, which is a very nice property.

My question is, for a general $$nN\times nN$$ matrix with $$N\times N$$ blocks where all of the blocks commute with each other, can we find the determinant in a similar way? That is, by first finding the determinant treating the blocks like scalars and then taking the determinant of the resulting $$N\times N$$ matrix.

• If $D$ is not invertible then what about this identity $\det(\mathbf{S})=\det(AD-BD^{-1}CD)$ – Sry Jun 20 '15 at 6:54