Inverse of a 3x3 block matrix

Question

I would like to get the inverse of a 3x3 (covariance) block matrix

begin{bmatrix}A&B&C\B'&D&E\C'&E'&Fend{bmatrix}

where the prime ' indicates the transposition operator.

Is there any general formula (or a way to solve this problem)?

Thank you very much.

MRule · Accepted Answer

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: $tilde A=E$, $tilde B=begin{bmatrix}F&Gend{bmatrix}$, $tilde C=begin{bmatrix}H\Lend{bmatrix}$, and $tilde D=begin{bmatrix}J&K\M&Nend{bmatrix}$:
$$
begin{aligned}
X^{-1} &=
begin{bmatrix}
E&begin{bmatrix}F&Gend{bmatrix}\
begin{bmatrix}H\Lend{bmatrix}&begin{bmatrix}J&K\M&Nend{bmatrix}
end{bmatrix}^{-1}
\&=
begin{bmatrix}
E^{-1}+E^{-1}begin{bmatrix}F&Gend{bmatrix}Z^{-1}begin{bmatrix}H\Lend{bmatrix}E^{-1}&-E^{-1}begin{bmatrix}F&Gend{bmatrix}Z^{-1}\
-Z^{-1}begin{bmatrix}H\Lend{bmatrix}E^{-1}&Z^{-1}
end{bmatrix}
\
Z &= begin{bmatrix}J&K\M&Nend{bmatrix} - begin{bmatrix}H\Lend{bmatrix} E^{-1} begin{bmatrix}F&Gend{bmatrix}
end{aligned}
$$
The factor $Z$, in turn, is another 2×2 block matrix.
$$
begin{aligned}
Z
&= 
begin{bmatrix}J&K\M&Nend{bmatrix} - begin{bmatrix}H\Lend{bmatrix} E^{-1} begin{bmatrix}F&Gend{bmatrix}
\
&= 
begin{bmatrix}J&K\M&Nend{bmatrix} - begin{bmatrix}
HE^{-1}F & HE^{-1}G
\
LE^{-1}F & LE^{-1}G
end{bmatrix} \
&= 
begin{bmatrix}
J-HE^{-1}F
&
K-HE^{-1}G
\
M-LE^{-1}F
&
N-LE^{-1}G
end{bmatrix} 
end{aligned}
$$
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:
$$
begin{aligned}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
E^{-1}&+E^{-1}begin{bmatrix}F&Gend{bmatrix}Z^{-1}begin{bmatrix}H\Lend{bmatrix}E^{-1}
\&=
E^{-1}left{I+
begin{bmatrix}F&Gend{bmatrix}
begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\-S^{-1}CA^{-1}&S^{-1}end{bmatrix}
begin{bmatrix}H\Lend{bmatrix}E^{-1}
right}
\&=
E^{-1}left{I+
begin{bmatrix}F&Gend{bmatrix}
begin{bmatrix}[A^{-1}+A^{-1}BS^{-1}CA^{-1}]H-A^{-1}BS^{-1}L\-S^{-1}CA^{-1}H + S^{-1}Lend{bmatrix}
E^{-1}
right}
\&=
E^{-1}left{I
+left{
Fleft[(A^{-1}+A^{-1}BS^{-1}CA^{-1})H-A^{-1}BS^{-1}Lright]
+Gleft[-S^{-1}CA^{-1}H + S^{-1}Lright]
right}E^{-1}
right}
\&=
E^{-1}left{I
+
left{
FA^{-1}left[H+BS^{-1}(CA^{-1}H-L)right]
-GS^{-1}[CA^{-1}H-L]
right}
E^{-1}
right}
\&=
E^{-1}left{I+left{FA^{-1}H+left[FA^{-1}B-Gright]S^{-1}[CA^{-1}H-L]right}E^{-1}right}
\&=
E^{-1}+E^{-1}left{FA^{-1}H+US^{-1}Vright}E^{-1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\
\
-&E^{-1}begin{bmatrix}F&Gend{bmatrix}Z^{-1}
\&=
-E^{-1}
begin{bmatrix}F&Gend{bmatrix}
begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\-S^{-1}CA^{-1}&S^{-1}end{bmatrix}
\&=
-E^{-1}begin{bmatrix}
FA^{-1}+FA^{-1}BS^{-1}CA^{-1}-GS^{-1}CA^{-1}
&
-FA^{-1}BS^{-1}+GS^{-1}
end{bmatrix}
\&=
begin{bmatrix}
-E^{-1}left{
F+(FA^{-1}B-G)S^{-1}C
right}
A^{-1}
&
E^{-1}[FA^{-1}B-G]S^{-1}
end{bmatrix}
\&=
begin{bmatrix}
-E^{-1}left[
F-US^{-1}C
right]
A^{-1}
&
-E^{-1}US^{-1}
end{bmatrix}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\
\
-&Z^{-1}begin{bmatrix}H\Lend{bmatrix}E^{-1}
\&=-begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\-S^{-1}CA^{-1}&S^{-1}end{bmatrix}
begin{bmatrix}H\Lend{bmatrix}E^{-1}
\&=
begin{bmatrix}
-A^{-1}H-A^{-1}BS^{-1}CA^{-1}H + A^{-1}BS^{-1}L
\
S^{-1}CA^{-1}H -S^{-1}L
end{bmatrix}E^{-1}
\&=
begin{bmatrix}
-A^{-1}[H+BS^{-1}(CA^{-1}H-L)]E^{-1}
\
S^{-1}(CA^{-1}H-L)E^{-1}
end{bmatrix}
\&=
begin{bmatrix}
-A^{-1}[H-BS^{-1}V]E^{-1}
\
-S^{-1}VE^{-1}
end{bmatrix}
end{aligned}
$$
This gives the expanded formula:
$$
begin{aligned}
&X^{-1}=
\&
begin{bmatrix}
E^{-1}+E^{-1}left[FA^{-1}H+US^{-1}Vright]E^{-1}
&
-E^{-1}left[F-US^{-1}Cright]A^{-1}
&
-E^{-1}US^{-1}
\
-A^{-1}[H-BS^{-1}V]E^{-1}
&
A^{-1}+A^{-1}BS^{-1}CA^{-1}
&
-A^{-1}BS^{-1}
\
-S^{-1}VE^{-1}
&
-S^{-1}CA^{-1}
&
S^{-1}
end{bmatrix}
end{aligned}
$$

Inverse of a 3x3 block matrix

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