Matrix Determinant Formula 
Problem. Let $A$ be a non-singular $n\times n$ matrix and let $\Gamma=[\Gamma_1\quad\Gamma_2]$ be an $n\times n$ orthogonal matrix where $\Gamma_1$ is $n\times n_1$, $\Gamma_2$ is $n\times n_2$ and $n=n_1+n_2$. Show that $$\det(\Gamma_1^TA\Gamma_1)=\det(A)\det(\Gamma_2^TA^{-1}\Gamma_2).$$

My Attempts. Here we make use of the property of orthogonal matrix:
\begin{align}
\det(A)=\det(\Gamma^TA\Gamma)=\det\left(\begin{bmatrix}
\Gamma_1^T \\ \Gamma_2^T
\end{bmatrix}A\begin{bmatrix}
\Gamma_1 & \Gamma_2
\end{bmatrix}\right)=\det\left(\begin{bmatrix}
\Gamma_1^TA\Gamma_1 & \Gamma_1^TA\Gamma_2 \\ \Gamma_2^TA\Gamma_1 & \Gamma_2^TA\Gamma_2
\end{bmatrix}\right). 
\end{align}
Since $A$ is non-singular, $\Gamma_1^TA\Gamma_1$ is also non-singular. Thus,
\begin{align}
\det(A)=\det(\Gamma_1^TA\Gamma_1)\det\left(\Gamma_2^TA\Gamma_2-\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right). 
\end{align}
If the formula we want to prove is true, we would have
\begin{align}
1&=\det(\Gamma_2^TA^{-1}\Gamma_2)\cdot\det\left(\Gamma_2^TA\Gamma_2-\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right) \\
&=\det\left(\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right). 
\end{align}
Nonetheless, I have no idea how to simplify the terms in the parenthesis because I only have $\Gamma_1\Gamma_1^T+\Gamma_2\Gamma_2^T=I$. Hope anyone has good suggestions.

Let's define $B = \Gamma^T A \Gamma$ and take the notation:
$$
\begin{aligned}
\Gamma_{11} = \Gamma_1^TA\Gamma_1 \\
\Gamma_{12} = \Gamma_1^TA\Gamma_2 \\
\Gamma_{21} = \Gamma_1 A \Gamma_2^T \\
\Gamma_{22} = \Gamma_2^T A\Gamma_2
\end{aligned}
$$
Let's denote the Shur complement of $\Gamma_{11}$ as $B/ \Gamma_{11} =  \Gamma_{2}^{T} A \Gamma_{2}-\Gamma_{2}^{T} A \Gamma_{1}\left(\Gamma_{1}^{T} A \Gamma_{1}\right)^{-1} \Gamma_{1}^{T} A \Gamma_{2}$ which is the same as $B / \Gamma_{11} = \Gamma_{22} - \Gamma_{21} \Gamma_{11}^{-1} \Gamma_{12}$. Using this notation we get:
$$
\begin{aligned}
&\det(A) = \det(B) = \det(\Gamma_{11})\det(B / \Gamma_{11})  \\
&\det(\Gamma_1^T A \Gamma_1) = \det(A)\det((B / \Gamma_{11})^{-1})
\end{aligned}
$$
We then compute $B^{-1}$:
$$ B^{-1} =
\left(
\begin{matrix}
\Gamma_{11}^{-1} + \Gamma_{11}^{-1} \Gamma_{12}(B / \Gamma_{11})^{-1}\Gamma_{21}\Gamma_{11}^{-1} & -\Gamma_{11}^{-1} \Gamma_{12}(B / \Gamma_{11})^{-1}\\
-(B / \Gamma_{11})^{-1} \Gamma_{21} \Gamma_{11}^{-1} & (B / \Gamma_{11} )^{-1}
\end{matrix}
\right)
$$
We also have
$$
B^{-1} = \Gamma^T A^{-1} \Gamma =
\left (
\begin{matrix}
\Gamma_1^T A^{-1} \Gamma_1 & \Gamma_1^T A^{-1} \Gamma_2 \\
\Gamma_1 A^{-1} \Gamma_2^T & \Gamma_2^T A^{-1} \Gamma_2
\end{matrix}
\right )
$$
By checking the dimensions we get:
$$
\Gamma_2^T A^{-1} \Gamma_2 = (B / \Gamma_{11} )^{-1}
$$
From where we get $\det(\Gamma_1^T A \Gamma_1) = \det(A)\det(\Gamma_2^T A^{-1} \Gamma_2 )$.
