Consider a matrix of the form $$ A:= \left[ \begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & 0 & 0 & a_{34}\\ 0 & a_{42} & a_{43} & 0 \end{array} \right] = \left[ \begin{array}{cc} A_{11} & A_{12}\\ A_{21} & A_{22} \end{array} \right], $$ where the entries $a_{ij}$, with $i,j=1,\ldots,4$, are complex functions and $A_{mn}$ are $2\times 2$ matrices. Is there a simpler way to compute $A^{-1}$ in this case? I've tried writing $A$ as a block matrix, and then using some ideas from the theory Schur complements to compute $A^{-1}$. However, this approach leads to monstrous entries that make the analytic computation of $A^{-1}$ very tough. Any ideas/suggestions are welcome.
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1This question notes the inverse of a $2 \times 2$ block matrix $$ S := \begin{bmatrix} A & B \ C & D \end{bmatrix} $$ is $$ S^{-1} = \begin{bmatrix} A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1} \end{bmatrix} $$ provided $A$ or $D$ are invertible. I'm not sure if this is too complicated for what you'd like though. – PrincessEev Sep 02 '24 at 07:02
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@PrincessEev That's exactly the formula that I used. Those entries are very tough to compute even with symbolic algebra packages like Maple. – user775349 Sep 02 '24 at 07:07
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@user775349: what makes you say "Those entries are very tough to compute"? What about the flexibility in partitioning the matrix e.g. into blocks of 2 x 2 or 3 x3? Are both partitions unusable for you? – Dennis Marx Sep 02 '24 at 14:09