Lagrange's identity is:
$$ \sum_{1\leq j <k\leq n}(a_jb_k-a_kb_j)^2=\left( \sum_{k=1}^na_k^2 \right)\left( \sum_{k=1}^n b_k^2 \right)-\left( \sum_{k=1}^na_kb_k \right)^2 $$
Why is the LHS sum such that it is? If one considers a $n\times n $ matrix, and the elements (j, k) in it, then reading the sum in the matrix shows that one picks values from the upper half above the diagonal.
Is there some insight to be gained from the matrix to understand why the Lagrange identity is such that it is?
Absolutely, Lagranges Identity can be interpreted geometrically as follows. I am not aware of your background so I will skip a proof which shows the following but you can expand the following expression given vectors $\vec a$ and $\vec b$ with $n$ components, the LHS of the identity is exactly:
$$(\vec a \times \vec b)\bullet(\vec a \times \vec b)$$
The funny-ness of it is also apparent when using some tricks in tensor algebra if you are interested. Pavel Grinfield's book on tensor calculus of moving surfaces is a great book to learn more about it.