 Recursive inverse of a lower triangular matrix

For an invertible lower triangular matrix, the scalar pivot $a$ and the remaining triangular block $C$ are both invertible. The displayed formula follows by matrix multiplication: the lower-left block in the product is $b/a-C(C^{-1}b/a)=0$, and the diagonal blocks are identities. Recursively invert $C$, compute $-C^{-1}b/a$, and assemble the inverse matrix; the one-dimensional base case is $a^{-1}$. This recursion reduces the size by one at each step.