TITLE: Matrix Determinant Identity
QUESTION [9 upvotes]: I have come across an observation about the determinant of a matrix, but I don't know how to prove it in general. Let me demonstrate it through an example.
$$
\begin{align}
\left|
\begin{matrix}
 1 & 2 & 3 & 4 \\
 5 & 6 & 7 & 8 \\
 9 & 10 & 11 & 12 \\
 13 & 14 & 15 & 16 \\
\end{matrix}
\right|
\cdot
\left|
\begin{matrix}
 6 & 7 \\
 10 & 11 \\
\end{matrix}
\right|
&=
\left|
\begin{matrix}
 1 & 2 & 3 \\
 5 & 6 & 7 \\
 9 & 10 & 11 \\
\end{matrix}
\right|
\cdot\left|
\begin{matrix}
 6 & 7 & 8 \\
 10 & 11 & 12 \\
 14 & 15 & 16 \\
\end{matrix}
\right|\\
&-
\left|
\begin{matrix}
 2 & 3 & 4 \\
 6 & 7 & 8 \\
 10 & 11 & 12 \\
\end{matrix}
\right|
\cdot
\left|
\begin{matrix}
 5 & 6 & 7 \\
 9 & 10 & 11 \\
 13 & 14 & 15 \\
\end{matrix}
\right|
\end{align}
$$
What this essentially says is that you can find the determinant of a larger matrix by breaking it down into this "ad-bc"-style of smaller determinants, however I can't work out why this should be true. I've expanded out the determinant for a 3x3 example in full generality and confirmed that it works, but I'd like to know if there's a more straightforward reason for this observation.

REPLY [2 votes]: This is a determinant identity for Schur complement of a matrix. It's also closely related to http://en.wikipedia.org/wiki/Dodgson_condensation. "It is named for its inventor Charles Dodgson (better known as Lewis Carroll)."