General Formula: Matrix Inversion Lemma

Let ${\bf A}$, ${\bf C}$, and ${\bf C}^{-1}+{\bf D}{\bf A}^{-1}{\bf B}$be non-singular square matrices; then

$$({\bf A}+{\bf B}{\bf C}{\bf D})^{-1} = {\bf A}^{-1}-{\bf A}^{... ...}+{\bf D}{\bf A}^{-1}{\bf B} \right)^{-1} {\bf D}{\bf A}^{-1}.$$

General Formula: Matrix Inversion in Block form

Let a $m \times n$ matrix ${\bf M}$ be partitioned into a block form:

$$\begin{array}{rcl} {\bf M}& = & \left[ \begin{array}{cc} {\bf... ...c} \underbrace{\;}_m \underbrace{\;}_n \end{array}\end{array}$$

where the $m \times m$ matrix ${\bf A}$ and $n \times n$ matrix ${\bf D}$ are invertible. Then we have

$$\left[ \begin{array}{cc} {\bf A}& {\bf B}\\ {\bf C}& {\bf D}\... ...& ({\bf D}-{\bf C}{\bf A}^{-1}{\bf B})^{-1} \end{array}\right]$$

$$\left[ \begin{array}{cc} {\bf X}& {\bf Y}\\ {\bf Z}& {\bf U}\... ...& ({\bf D}-{\bf C}{\bf A}^{-1}{\bf B})^{-1} \end{array}\right]$$

It can be proved that the above two matrix expressions for ${\bf M}^{-1}$are equivalent.

Special Case 1

Let a $(m+1) \times (m+1)$ matrix ${\bf M}$ be partitioned into a block form:

$$\begin{array}{rcl} {\bf M}& = & \left[ \begin{array}{cc} {\bf... ...c} \underbrace{\;}_m \underbrace{\;}_1 \end{array}\end{array}$$

Then the inverse of ${\bf M}$ is

$${\bf M}^{-1} = \left[ \begin{array}{cc} \left( {\bf A}- \frac... ...{-1} & \frac{\textstyle 1}{\textstyle k}\\ \end{array}\right]$$

where $k=c-{\bf b}^T{\bf A}^{-1}{\bf b}$.

Special Case 2

Suppose that we have a given matrix equation

$$\left[ \begin{array}{cc} {\bf A}& {\bf B}\\ {\bf C}& {\bf D}... ...t[ \begin{array}{c} {\bf I}\\ \mbox{\bf0} \end{array}\right],$$ (1)

where ${\bf A}$ and ${\bf D}$ are invertible matrices and all matrices are of compatible dimensions in the above equation. Please find the matrices ${\bf X}$ and ${\bf Y}$ in terms of the given matrices ${\bf A}$, ${\bf B}$, ${\bf C}$, and ${\bf D}$.