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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.2012-10-31
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    Remark on that comment: That follows from [Cramer's Rule](https://en.wikipedia.org/wiki/Cramer%27s_rule#Finding_inverse_matrix). (@IHaveAStupidQuestion)2012-10-31
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    @K.Stm. I know, that was my point.2012-10-31
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    @IHaveAStupidQuestion Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-08-22

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