Now, to exhibit the full factorization A = LU, we need to compute the
product
L = L1-1
L2-1 L3-1 .
Perhaps surprisingly, this turns out to be a triviality. The inverse
of L1 is just L1 itself, but with each entry
below the diagonal negated
The inverses of L2 and L3 are obtained similarly: just negate their sub diagonal entries. Finally, the product L1-1 L2-1 L3-1 is just the unit lower-triangular matrix with the nonzero sub diagonal entries of L1-1, L2-1, and L3-1 inserted in the appropriate places. Thus