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Let $a,b,c$ be reals such that $ab-c^2\neq0$, and let $$ A = \begin{pmatrix} a & 0 & c & 0 \\ 0 & a & 0 & c \\ c & 0 & b & 0 \\ 0 & c & 0 & b \end{pmatrix}. $$ The matrix inverse is $$ A^{-1} = \frac1{ab-c^2}\begin{pmatrix} b & 0 & -c & 0 \\ 0 & b & 0 & -c \\ -c & 0 & a & 0 \\ 0 & -c & 0 & a \end{pmatrix}. $$ This very curiously (in my opinion) resembles the formula of the $2\times 2$ inverse matrix applied to the $2 \times 2$ diagonal blocks composing the matrix.

What's the rule I'm not seeing?

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You can get this immediately via abuse of notation:

$$ A = \begin{pmatrix} a & 0 & c & 0 \\ 0 & a & 0 & c \\ c & 0 & b & 0 \\ 0 & c & 0 & b \end{pmatrix} = \begin{pmatrix} a 𝟙 & c 𝟙 \\ c 𝟙 & b 𝟙\end{pmatrix} $$

Now you have a $2\times 2$ matrix and can apply the usual formula.

If you want it a bit more formal, use $A= \begin{pmatrix}a & c \\ c & b\end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\left( M \otimes N \right)^{-1} = M^{-1} \otimes N^{-1}$

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