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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 $D$?

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up vote 3 down vote accepted

Take a look at: or 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. – user17762 May 20 '12 at 4:22

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