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.

To simplify the terms, we indeed have to use the formula $\Gamma_2\Gamma_2^T=I-\Gamma_1\Gamma_1^T$. Here
\begin{align*}
\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_2&=\Gamma_2^TA^{-1}(I-\Gamma_1\Gamma_1^T)A\Gamma_2 \\
&=\Gamma_2^TA^{-1}A\Gamma_2-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1A\Gamma_2 \\
&=I-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1A\Gamma_2
\end{align*}
and
\begin{align*}
&\quad\ \Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\
&=\Gamma_2^TA^{-1}(I-\Gamma_1\Gamma_1^T)A\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\
&=\Gamma_2^TA^{-1}A\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\
&=\color{red}{\Gamma_2^T\Gamma_1}(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_1\color{red}{\Gamma_1^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}}\Gamma_1^TA\Gamma_2 \\
&=0-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_2=-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_2. 
\end{align*}
Thus,
\begin{align*}
&\quad\ \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 \\
&=I-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1A\Gamma_2-(-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_2)=I. 
\end{align*}
The formula thus follows.
P.S. If anyone has simpler solutions, please share it with me!! This calculation is too complicated.
