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Let $A$ be a symmetric and invertible matrix. I know that if $A$ has a constant diagonal and a constant off-diagonal ($A=\alpha I + \beta\tilde1$, where $I$ is an identity matrix, $\tilde{1}$ is a matrix of ones, and $\alpha,\beta$ are some scalars), then $A^{-1}$ has a constant diagonal.

What are other non-obvious possible structures of $A$ that guarantee that $A^{-1}$ has a constant diagonal vector?

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$A^{-1}$ has constant diagonal if and only if there exists a constant $C$ such that $\det(A_{ii})=C$ for all $i$ where $A_{ii}$ denotes the matrix with $i$th row and $i$th column deleted. – IHaveAStupidQuestion Oct 31 '12 at 20:16
Remark on that comment: That follows from Cramer's Rule. (@IHaveAStupidQuestion) – K. Stm. Oct 31 '12 at 20:19
@K.Stm. I know, that was my point. – IHaveAStupidQuestion Oct 31 '12 at 20:22

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