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Considering a matrix given by composition of square matrices like:

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

I want to calculate its determinant $|M|$. Consider that all components are square matrices of course.

Can this be related somehow to the components $A$, $B$, $C$ and $D$?

2 Answers 2

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Take a look at: http://arxiv.org/abs/1112.4379 or http://www.hostemostel.com/pdf/MTH/0.pdf for a detailed answer. In particular, for your question, if $A,B,C,D$ commute with each other: $\det M = \det(AD-BC)$.

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Note that $\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \begin{pmatrix} A & 0 \\ C & I \end{pmatrix} \begin{pmatrix} I & A^{-1}B \\ 0 & D - CA^{-1}B \end{pmatrix} = \begin{pmatrix} A - BD^{-1}C & BD^{-1} \\ 0 & I \end{pmatrix} \begin{pmatrix} I & 0 \\ C & D\end{pmatrix}$ The middle factorization is valid assuming $A$ is invertible and the last factorization is valid assuming $D$ is invertible. Further, $\det \left( \begin{pmatrix} X_{11} & X_{12}\\ 0 & X_{22} \end{pmatrix} \right) = \det(X_{11}) \det(X_{22})$ and $\det \left( \begin{pmatrix} X_{11} & 0\\ X_{21} & X_{22} \end{pmatrix} \right) = \det(X_{11}) \det(X_{22})$ Hence, $\det \left( \begin{pmatrix} A & B \\ C & D \end{pmatrix} \right) = \det(A) \det(D-CA^{-1}B) = \det(A-BD^{-1}C) \det(D)$ The middle one assumes $A$ is invertible, and the last one assumes $D$ is invertible.

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    @JasonDeVito:$ $Yes.$ $Thanks for pointing it out. Will add it.2012-05-20