A Problem on Linear Algebra I'm trying to calculate an integral over the generalized Poincare upper half plane, then I find that I need to show the following identity:  

Let $X=(X_{i,j})\in\mathrm{GL}(n,\mathbb R)(n\geq 3)$ and $r$ be a positive integer such that $2\leq r\leq n-1$. For any $1\leq \ell_1<\ell_2<\cdots< \ell_r\leq n$, we define
  \begin{eqnarray*}
\det(\ell_1,\ell_2,\cdots,\ell_r):=\det\left\{
\left(
  \begin{array}{cccc}
    X_{n-r+1,\ell_1} & X_{n-r+1,\ell_2} & \cdots & X_{n-r+1,\ell_r} \\
    X_{n-r+2,\ell_1} & X_{n-r+2,\ell_2} & \cdots & X_{n-r+2,\ell_r} \\
    \vdots & \vdots & \cdots & \vdots \\
    X_{n,\ell_1} & X_{n,\ell_2} & \cdots & X_{n,\ell_r} \\
  \end{array}
\right)\right\}.
\end{eqnarray*}
  Let 
  \begin{eqnarray*}
A&=&\left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r-1}\leq n-1}[\det(\ell_1,\cdots,\ell_{r-1})]^2\right)\times\left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r-1}\leq n-1}[\det(\ell_1,\cdots,\ell_{r-1},n)]^2\right)\\
B&=&\left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r}\leq n-1}[\det(\ell_1,\cdots,\ell_{r})]^2\right)\cdot\left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r-2}\leq n-1}[\det(\ell_1,\cdots,\ell_{r-2},n)]^2\right)\\
C&=&  \left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r-1}\leq n-1}\det(\ell_1,\cdots,\ell_{r-1})\cdot \det(\ell_1,\cdots,\ell_{r-1},n)\right)^2.
\end{eqnarray*}
  Show that $A-B=C$.

The cases $n=3,4$ can be verified by Mathematical.
The case $r=2$ with any $n\geq 3$ can be verified easily by a direct calculation, but how to prove this identity for general $r$ ? Any comments and suggestions are welcomed!
 A: The following proof is due to Yeping ZHANG:
Let $V$ be a $n$ dimensional Euclidean vector space. Let $(e_1,\cdots,e_n)$ be an orthogonal basis.
For any $k=1,\cdots,n$, let $\Lambda^k V$ the $k$-th exterior product of $V$. We equip $\Lambda^k V$ with a metric ${\lVert \cdot \rVert}_{\Lambda^k V}$ such that $(e_{l_1}\wedge\cdots\wedge e_{l_k})_{1\leq l_1 < \cdots < l_k\leq n}$ is an orthogonal basis. Let ${\langle \cdot,\cdot \rangle}_{\Lambda^k V}$ be the associated inner product.
Let $v_1,\cdots,v_k\in V$, whose coodinates with respect to $(e_1,\cdots,e_n)$ are given by
\begin{equation}
v_i = \sum_{j=1}^n t_{ij}e_j \;,
\end{equation}
where $i = 1,\cdots,k$. Let $T=(t_{ij})_{1\leq i \leq k, 1\leq j \leq n}$ be the associated matrix. For $1\leq l_1 < \cdots < l_k\leq n$, set $T(l_1,\cdots,l_k)=(t_{ij})_{1\leq i \leq k, j = l_1,\cdots l_k}$. Then
\begin{equation}
v_1 \wedge\cdots\wedge v_k = \sum_{1\leq l_1 < \cdots < l_k\leq n} \, \det \, T(l_1,\cdots,l_k) \, e_{l_1}\wedge\cdots\wedge e_{l_k} \;,
\end{equation}
in particular
\begin{equation}
{\lVert v_1 \wedge\cdots\wedge v_k \rVert}_{\Lambda^k V}^2 = \sum_{1\leq l_1 < \cdots < l_k\leq n} [ \, \det \, T(l_1,\cdots,l_k) \, ]^2 \;.
\end{equation}
With the above observation, we can interpret the question as the follows.
Set
\begin{equation}
x_i = \sum_{j=1}^n X_{ij}e_j \;,
\end{equation}
where $i=1,\cdots,n$. Let
\begin{align}
x_{n-r+2} \wedge\cdots\wedge x_n = \alpha + \beta \wedge e_n \;,\nonumber\\
x_{n-r+1} \wedge\cdots\wedge x_n = \gamma + \delta \wedge e_n \;,
\end{align}
with $\alpha,\beta,\gamma,\delta$ generated by $e_1,\cdots,e_{n-1}$.
We have
\begin{align}
A & = {\lVert \alpha \rVert}_{\Lambda^{r-1} V}^2 \cdot {\lVert \delta \rVert}_{\Lambda^{r-1} V}^2 \;,\nonumber\\
B & = {\lVert \gamma \rVert}_{\Lambda^r V}^2 \cdot {\lVert \beta \rVert}_{\Lambda^{r-2} V}^2 \;,\nonumber\\
C & = {\langle \alpha , \delta \rangle}_{\Lambda^{r-1} V}^2\;.
\end{align}
We have
\begin{equation}
x_i = y_i + X_{i,n}e_n \;,
\end{equation}
where $y_i = X_{i,1}e_1 + \cdots X_{i,n-1}e_{n-1}$. Then
\begin{align}
\alpha & = y_{n-r+2}\wedge\cdots\wedge y_n \;,\nonumber\\
\beta & = \sum_{l=n-r+2}^n (-1)^{n-l} X_{l,n} y_{n-r+2}\wedge\cdots\wedge y_{l-1}\wedge y_{l+1}\wedge\cdots\wedge y_n \;,\nonumber\\
\gamma & =  y_{n-r+1}\wedge\cdots\wedge y_n \;,\nonumber\\
\delta & = \sum_{l=n-r+1}^n (-1)^{n-l} X_{l,n} y_{n-r+1}\wedge\cdots\wedge y_{l-1}\wedge y_{l+1}\wedge\cdots\wedge y_n \;.
\end{align}
Case $1$. If one of $y_{n-r+2},\cdots,y_n$ is zero, we have
\begin{equation}
\alpha = \gamma = 0 \;,
\end{equation}
trivially, we get $A=B+C$.
Case 2. If $y_{n-r+1}=0$, we have
\begin{align}
\gamma & = 0 \;,\nonumber\\
\delta & = (-1)^{r-1}X_{n-r+1,n}\alpha \;.
\end{align}
Thus
\begin{align}
A & = X_{n-r+1,n}^2 {\lVert \alpha \rVert}_{\Lambda^{r-1} V}^4 \;,\nonumber\\
B & = 0 \;,\nonumber\\
C & = X_{n-r+1,n}^2 {\lVert \alpha \rVert}_{\Lambda^{r-1} V}^4 \;.
\end{align}
We find $A=B+C$.
Case 3. We suppose that none of $y_{n-r+1},\cdots,y_n$ is zero.
Without changing $\alpha,\beta,\gamma,\delta$, we can replace $x_i$ by $x_i - tx_j$ for any $j>i$, any $t\in\mathbb{R}$. Thus we can suppose that
\begin{equation}
\langle y_i,y_j \rangle = 0 \;,
\end{equation}
for any $i \neq j$. Since $A,B,C$ are homogenious functions of $y_{n-r+1},\cdots,y_n$, we can suppose that $\lVert y_i \rVert = 1$.
With the above assumptions, we have
\begin{align}
& {\lVert \alpha \rVert}_{\Lambda^{r-1} V}^2 = 1 \;,\nonumber\\
& {\lVert \beta \rVert}_{\Lambda^{r-2} V}^2 = \sum_{l=n-r+2}^n X_{l,n}^2 \;,\nonumber\\
& {\lVert \gamma \rVert}_{\Lambda^{r} V}^2 = 1 \;,\nonumber\\
& {\lVert \delta \rVert}_{\Lambda^{r-1} V}^2 = \sum_{l=n-r+1}^n X_{l,n}^2 \;,\nonumber\\
& {\langle \alpha , \delta \rangle}_{\Lambda^{r-1} V}^2 = X_{n-r+1,n}^2 \;.
\end{align}
Hence $A = B + C$.
