# proof of block determinants

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

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

So

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

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 ProofOfBlockDeterminants 2013-03-22 15:27:49 2013-03-22 15:27:49 georgiosl (7242) georgiosl (7242) 5 georgiosl (7242) Proof msc 15A15