Binet-Cauchy Identity/Proof 2

Proof
This is a special case of the Cauchy-Binet Formula:
 * $\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m}}\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$

where:
 * $\mathbf A$ is an $m \times n$ matrix
 * $\mathbf B$ is an $n \times m$ matrix.


 * For $1 \le j_1, j_2, \ldots, j_m \le n$:
 * $\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.


 * $\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.

In this case $m = 2$, giving:
 * $\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop \le n} \det \left({\mathbf A_{j_1 j_2} }\right) \det \left({\mathbf B_{j_1 j_2} }\right)$