TITLE: The inverse of a block-upper triangular matrix
QUESTION [13 upvotes]: 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 ?

REPLY [15 votes]: 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 $0$.