# Inverse of 2 by 2 block matrix

Suppose

$$R = \begin{bmatrix} A & B\\ C & D\end{bmatrix}$$

is a $$2 \times 2$$ block matrix of real numbers, where $$A$$ and $$D$$ are squared diagonal matrices.

Is it possible that the following four conditions hold simultaneously?

1. $$R$$ is invertible

2. $$D$$ is nonsingular

3. the Schur complement of $$D$$, $$A-BD^{-1}C$$ is singular.

4. $$A$$ is singular.

If so, could you please provide a way to find the inverse $$R$$ in terms of the partitions of $$R$$?

$$\begin{pmatrix}A&B\\C&D\end{pmatrix}^{-1}=\begin{pmatrix}\left(A-B D^{-1}C\right)^{-1}&-\left(A-B D^{-1}C\right)^{-1} BD^{-1}\\-D^{-1}C\left(A-B D^{-1}C\right)^{-1}&D^{-1}+D^{-1}C\left(A-BD^{-1}C \right)^{-1}BD^{-1}\end{pmatrix}$$
• The formula you mention only holds if $D$ and $A-BD^{-1}C$ are invertible. However, I'm curious what will happen if $D$ is invertible but $A-BD^{-1}C$ and $A$ are not invertible? Commented Sep 13, 2022 at 4:33
Suppose $$A$$ is a $$1\times 1$$ equal to zero. Let $$D$$ be any invertible matrix and $$C$$ any column vector. Choose $$B$$ to be a row vector which is orthogonal to $$D^{-1}C.\;$$ These choices will meet all of the stated conditions.
Assume that the initial elements of $$\{B,C,D\}$$ are $$\{1,2,3\},$$ respectively. Then repartitioning produces $$A=\pmatrix{0&1\\2&3}$$ or some other invertible matrix depending on those initial elements.
The important point is that the invertibility of new $$A$$ means that $$R^{-1}$$ can be calculated using its Schur complement.