= Recursive inverse of a lower triangular matrix
{title2=$\begin{pmatrix}a&0\\b&C\end{pmatrix}^{-1}=\begin{pmatrix}a^{-1}&0\\-C^{-1}b/a&C^{-1}\end{pmatrix}$}
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.
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