Now, to exhibit the full factorization A = LU, we need to compute the
product

L = L_{1}^{-1}
L_{2}^{-1} L_{3}^{-1} .
Perhaps surprisingly, this turns out to be a triviality. The inverse
of L_{1} is just L_{1} itself, but with each entry
below the diagonal negated

The inverses of L_{2} and L_{3} are obtained
similarly: just negate their sub diagonal entries. Finally, the
product
L_{1}^{-1}
L_{2}^{-1} L_{3}^{-1} is just
the unit lower-triangular matrix with the nonzero sub diagonal entries
of
L_{1}^{-1},
L_{2}^{-1}, and L_{3}^{-1}
inserted in the appropriate places. Thus