Problem: Let $M$ be an invertible matrix of dimension $2n \times 2n$, represented in block form as
\[
M = \begin{bmatrix}
A & B \\
C & D \\
\end{bmatrix}, \kern 0.5cm
\text{ and } \kern 0.5cm
\begin{bmatrix}
E & F \\
G & H \\
\end{bmatrix}.
\]
Then prove that
\[
\det M \cdot \det I = - \det C.
\]
Solution: Let $I$ denotes the identity matrix. Observe that
\[
\begin{bmatrix}
A & B \\
C & D \\
\end{bmatrix} \begin{bmatrix}
E & F \\
G & H \\
\end{bmatrix} = \begin{bmatrix}
I & 0 \\
0 & I \\
\end{bmatrix} \implies \begin{bmatrix}
AE + BG & AF + BH \\
CE + DG & CF + DH \\
\end{bmatrix} =
\begin{bmatrix}
I & 0 \\
0 & I \\
\end{bmatrix}.
\]
Note that
\[
\det
\begin{bmatrix}
E & I \\
G & 0 \\
\end{bmatrix} = -\det G.
\]
Therefore,
\begin{align*}
\det M \cdot (-\det G) & = \det \begin{bmatrix}
A & B \\
C & D \\
\end{bmatrix} \cdot \det \begin{bmatrix}
E & I \\
G & 0
\end{bmatrix} \\
& = \det \begin{bmatrix}
A & B \\
C & D \\
\end{bmatrix}
\begin{bmatrix}
E & I \\
G & 0
\end{bmatrix} \\
& = \det
\begin{bmatrix}
AE + BG & A \\
CE + DG & C
\end{bmatrix} \\
& = \det
\begin{bmatrix}
I & A \\
0 & C \\
\end{bmatrix} \\
& = \det C.
\end{align*}
Therefore, we proved that
\[
\det M \cdot \det G = -\det C.
\]