proof of block determinants

If $A^{-1}$ exists, then

\begin{pmatrix}A&B\\ C&D\end{pmatrix}=\begin{pmatrix}I&O\\ CA^{-1}&I\end{pmatrix}\begin{pmatrix}A&B\\ O&D-CA^{-1}B\end{pmatrix}.

\det\begin{pmatrix}A&B\\ C&D\end{pmatrix}=\det\begin{pmatrix}I&O\\ CA^{-1}&I\end{pmatrix}\det\begin{pmatrix}A&B\\ O&D-CA^{-1}B\end{pmatrix}.

Each of the first matrices in the decompositions are triangular. Hence their determinants equal $1$ . This means that the determinant of the original matrix equals the determinant of either of the second matrices in the decomposition. Therefore

\det\begin{pmatrix}A&B\\ C&D\end{pmatrix}=\det(A)\det(D-CA^{-1}B).

The second formula follows by using a similar trick.

Title	proof of block determinants
Canonical name	ProofOfBlockDeterminants
Date of creation	2013-03-22 15:27:49
Last modified on	2013-03-22 15:27:49
Owner	georgiosl (7242)
Last modified by	georgiosl (7242)
Numerical id	5
Author	georgiosl (7242)
Entry type	Proof
Classification	msc 15A15