# How to derive a certain determinantal identity

Does anybody know how to derive

$$\det(\mathbf A)\cdot \det(\mathbf D + \mathbf E \cdot \mathbf A^{-1} \cdot \mathbf B) = \det(\mathbf D)\cdot \det(\mathbf A + \mathbf B \cdot \mathbf D^{-1} \cdot \mathbf E)$$

(most likely using Sylvester determinant theorem), where $\mathbf A$,$\mathbf B$,$\mathbf C$,$\mathbf D$,$\mathbf E$ are non-singular matrices?

UPDATE: corrected

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I don't think this can be true: the right side is $\det D \det B \det D^{-1} \det E = \det B \det E$, which is independent of $D$. If we take $D = -EA^{-1}B$ then the left side is 0. – Eric O. Korman Dec 11 '10 at 14:47
I must have made a mistake with putting the formula in the form, now the correct one in the question body – matcheek Dec 11 '10 at 14:56
also thanks for putting that in latex – matcheek Dec 11 '10 at 14:56
You can prove it just by using $\det{X\cdot Y}=\det{X}\det{Y}$ and usual multiplication rules for matrices. – Raskolnikov Dec 11 '10 at 15:02
One might get started with the special case $A=D=I$, proving: $$det(I+EB)=det(I+BE)$$ for any invertible matrices $B,E$. – hardmath Dec 11 '10 at 16:00

\begin{align} \det(A)\cdot\det(D+E\cdot A^{-1}\cdot B) & = \det(A)\cdot\det(D)\cdot\det(1+D^{-1}\cdot E\cdot A^{-1}\cdot B) \\ & = \det(A)\cdot\det(D)\cdot\det(B)\cdot\det(B^{-1}+D^{-1}\cdot E\cdot A^{-1}) \\ & = \det(A)\cdot\det(D)\cdot\det(1+B\cdot D^{-1}\cdot E\cdot A^{-1}) \\ & = \det(D)\cdot\det(A+B\cdot D^{-1}\cdot E) \end{align}
A neat way to prove this is by observing that these are two ways to write the determinant of the block matrix $$\pmatrix{ A & B \\ -E & D }$$ (see http://en.wikipedia.org/wiki/Determinant#Block_matrices)
The proof using Sylvester's determinant theorem is done by simply plugging in the right "pairs" of matrices into the right spots. That is, given the theorem as $$\det(I_p + XY ) = \det(I_n + YX)$$ Just let $X = EA^{-1}$ and $Y = BD^{-1}$. From there I think you can figure it out on your own.