The inverse of a block-upper triangular matrix
Is it true that $$\begin{pmatrix} A & * \\ 0 & B \\ \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & * \\ 0 & B^{-1} \\ \end{pmatrix}$$
where $A$ and $B$ are $m \times m$ and $n \times n$ invertible, and * is for unspecified blocks ?
Solution 1:
Just try and block multiply: $$ \begin{pmatrix} A & X \\ 0 & B \end{pmatrix} \begin{pmatrix} A^{-1} & Y \\ 0 & B^{-1} \end{pmatrix}= \begin{pmatrix} I & AY+XB^{-1} \\ 0 & I \end{pmatrix} $$ so we need $$ Y=-A^{-1}XB^{-1} $$ and the upper right corner is $0$.