TITLE: Inverse of a 3x3 block matrix
QUESTION [8 upvotes]: I would like to get the inverse of a 3x3 (covariance) block matrix
\begin{bmatrix}A&B&C\\B'&D&E\\C'&E'&F\end{bmatrix}
where the prime ' indicates the transposition operator.
Is there any general formula (or a way to solve this problem)?
Thank you very much.

REPLY [8 votes]: As user3556214 points out, one can apply the formula for inverting a 2×2 block matrix repeatedly, but it does not lead to nice results.
Recall the formula for the inverse of a 2×2 block matrix:
$$
\begin{aligned}
\begin{bmatrix}
A&B\\
C&D
\end{bmatrix}^{-1}
&=
\begin{bmatrix}
A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\
-S^{-1}CA^{-1}&S^{-1}
\end{bmatrix}
\\
S &= D - C A^{-1} B
\end{aligned}
$$
Now consider a 3×3 block matrix
$$
\begin{aligned}
X &=
\begin{bmatrix}
E&F&G\\
H&J&K\\
L&M&N
\end{bmatrix}
\end{aligned}
$$
Apply the 2×2 block inverse formula, plugging in: $\tilde A=E$, $\tilde B=\begin{bmatrix}F&G\end{bmatrix}$, $\tilde C=\begin{bmatrix}H\\L\end{bmatrix}$, and $\tilde D=\begin{bmatrix}J&K\\M&N\end{bmatrix}$:
$$
\begin{aligned}
X^{-1} &=
\begin{bmatrix}
E&\begin{bmatrix}F&G\end{bmatrix}\\
\begin{bmatrix}H\\L\end{bmatrix}&\begin{bmatrix}J&K\\M&N\end{bmatrix}
\end{bmatrix}^{-1}
\\&=
\begin{bmatrix}
E^{-1}+E^{-1}\begin{bmatrix}F&G\end{bmatrix}Z^{-1}\begin{bmatrix}H\\L\end{bmatrix}E^{-1}&-E^{-1}\begin{bmatrix}F&G\end{bmatrix}Z^{-1}\\
-Z^{-1}\begin{bmatrix}H\\L\end{bmatrix}E^{-1}&Z^{-1}
\end{bmatrix}
\\
Z &= \begin{bmatrix}J&K\\M&N\end{bmatrix} - \begin{bmatrix}H\\L\end{bmatrix} E^{-1} \begin{bmatrix}F&G\end{bmatrix}
\end{aligned}
$$
The factor $Z$, in turn, is another 2×2 block matrix.
$$
\begin{aligned}
Z
&= 
\begin{bmatrix}J&K\\M&N\end{bmatrix} - \begin{bmatrix}H\\L\end{bmatrix} E^{-1} \begin{bmatrix}F&G\end{bmatrix}
\\
&= 
\begin{bmatrix}J&K\\M&N\end{bmatrix} - \begin{bmatrix}
HE^{-1}F & HE^{-1}G
\\
LE^{-1}F & LE^{-1}G
\end{bmatrix} \\
&= 
\begin{bmatrix}
J-HE^{-1}F
&
K-HE^{-1}G
\\
M-LE^{-1}F
&
N-LE^{-1}G
\end{bmatrix} 
\end{aligned}
$$
Again apply again the 2×2 block inverse formula to get $Z^{-1}$, defining:
$$
\begin{aligned}
A &= J-HE^{-1}F
\\
B &= K-HE^{-1}G
\\
C &= M-LE^{-1}F
\\
D &= N-LE^{-1}G
\end{aligned}
$$
$$
\begin{aligned}
Z^{-1} &= 
\begin{bmatrix}
A&B\\
C&D
\end{bmatrix}^{-1}
=
\begin{bmatrix}
A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\
-S^{-1}CA^{-1}&S^{-1}
\end{bmatrix}
\\
S &= D - C A^{-1} B
\end{aligned}
$$
Further expanding the forumula for $X^{-1}$ in terms of this is tedious and somewhat unsatisfying. Define
$$
\begin{aligned}
U &= G - FA^{-1}B
\\
V &= L-CA^{-1}H
\end{aligned}
$$
then expand and simplify:
$$
\begin{aligned}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
E^{-1}&+E^{-1}\begin{bmatrix}F&G\end{bmatrix}Z^{-1}\begin{bmatrix}H\\L\end{bmatrix}E^{-1}
\\&=
E^{-1}\left\{I+
\begin{bmatrix}F&G\end{bmatrix}
\begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{bmatrix}
\begin{bmatrix}H\\L\end{bmatrix}E^{-1}
\right\}
\\&=
E^{-1}\left\{I+
\begin{bmatrix}F&G\end{bmatrix}
\begin{bmatrix}[A^{-1}+A^{-1}BS^{-1}CA^{-1}]H-A^{-1}BS^{-1}L\\-S^{-1}CA^{-1}H + S^{-1}L\end{bmatrix}
E^{-1}
\right\}
\\&=
E^{-1}\left\{I
+\left\{
F\left[(A^{-1}+A^{-1}BS^{-1}CA^{-1})H-A^{-1}BS^{-1}L\right]
+G\left[-S^{-1}CA^{-1}H + S^{-1}L\right]
\right\}E^{-1}
\right\}
\\&=
E^{-1}\left\{I
+
\left\{
FA^{-1}\left[H+BS^{-1}(CA^{-1}H-L)\right]
-GS^{-1}[CA^{-1}H-L]
\right\}
E^{-1}
\right\}
\\&=
E^{-1}\left\{I+\left\{FA^{-1}H+\left[FA^{-1}B-G\right]S^{-1}[CA^{-1}H-L]\right\}E^{-1}\right\}
\\&=
E^{-1}+E^{-1}\left\{FA^{-1}H+US^{-1}V\right\}E^{-1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\
\\
-&E^{-1}\begin{bmatrix}F&G\end{bmatrix}Z^{-1}
\\&=
-E^{-1}
\begin{bmatrix}F&G\end{bmatrix}
\begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{bmatrix}
\\&=
-E^{-1}\begin{bmatrix}
FA^{-1}+FA^{-1}BS^{-1}CA^{-1}-GS^{-1}CA^{-1}
&
-FA^{-1}BS^{-1}+GS^{-1}
\end{bmatrix}
\\&=
\begin{bmatrix}
-E^{-1}\left\{
F+(FA^{-1}B-G)S^{-1}C
\right\}
A^{-1}
&
E^{-1}[FA^{-1}B-G]S^{-1}
\end{bmatrix}
\\&=
\begin{bmatrix}
-E^{-1}\left[
F-US^{-1}C
\right]
A^{-1}
&
-E^{-1}US^{-1}
\end{bmatrix}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\
\\
-&Z^{-1}\begin{bmatrix}H\\L\end{bmatrix}E^{-1}
\\&=-\begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{bmatrix}
\begin{bmatrix}H\\L\end{bmatrix}E^{-1}
\\&=
\begin{bmatrix}
-A^{-1}H-A^{-1}BS^{-1}CA^{-1}H + A^{-1}BS^{-1}L
\\
S^{-1}CA^{-1}H -S^{-1}L
\end{bmatrix}E^{-1}
\\&=
\begin{bmatrix}
-A^{-1}[H+BS^{-1}(CA^{-1}H-L)]E^{-1}
\\
S^{-1}(CA^{-1}H-L)E^{-1}
\end{bmatrix}
\\&=
\begin{bmatrix}
-A^{-1}[H-BS^{-1}V]E^{-1}
\\
-S^{-1}VE^{-1}
\end{bmatrix}
\end{aligned}
$$
This gives the expanded formula:
$$
\begin{aligned}
&X^{-1}=
\\&
\begin{bmatrix}
E^{-1}+E^{-1}\left[FA^{-1}H+US^{-1}V\right]E^{-1}
&
-E^{-1}\left[F-US^{-1}C\right]A^{-1}
&
-E^{-1}US^{-1}
\\
-A^{-1}[H-BS^{-1}V]E^{-1}
&
A^{-1}+A^{-1}BS^{-1}CA^{-1}
&
-A^{-1}BS^{-1}
\\
-S^{-1}VE^{-1}
&
-S^{-1}CA^{-1}
&
S^{-1}
\end{bmatrix}
\end{aligned}
$$