Inverse of a 3×3 block matrix

 

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:

$$ \begin{aligned} &X^{-1}= \\& \begin{bmatrix} E^{-1}+E^{-1}\left[FA^{-1}H+US^{-1}V\right]E^{-1} & -E^{-1}\left[F-US^{-1}C\right]A^{-1} & -E^{-1}US^{-1} \\ -A^{-1}[H-BS^{-1}V]E^{-1} & A^{-1}+A^{-1}BS^{-1}CA^{-1} & -A^{-1}BS^{-1} \\ -S^{-1}VE^{-1} & -S^{-1}CA^{-1} & S^{-1} \end{bmatrix} \end{aligned} $$