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Basically, I have a $n \times n$ symmetric matrix, which looks like this:

$$ \begin{bmatrix} 1 & \alpha & \cdots & \alpha \\ \alpha & 1 & \cdots &\alpha \\ \vdots &\vdots & \ddots & \vdots \\ \alpha & \cdots & \alpha & 1 \end{bmatrix} $$

I would like to know is there a general formula / solution for solving the inverse of this matrix for $n = 1,2,3,4,5,6......$

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    $\begingroup$ How about if you carefully do 1 by 1, then 2 by 2, then 3 by 3, and if you still don't know, get back to us. $\endgroup$ – Will Jagy Apr 20 '14 at 22:57
  • $\begingroup$ I assume $\alpha\neq 1$, right? $\endgroup$ – Spock Apr 20 '14 at 22:58
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    $\begingroup$ The eigenvalues of the matrix are $\lambda_1=1+(n-1)\alpha$ and $\lambda_2=\cdots=\lambda_n=1-\alpha$. $\endgroup$ – mathse Apr 20 '14 at 23:04
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If $J$ is the $n \times n$ matrix of ones, then you want to find $B$ such that $$ \alpha BJ + (1- \alpha) B = I \tag{1} $$ Since $J^2 = nJ$, this implies $$ n \alpha B J + (1 - \alpha) B J = J $$ Therefore we must assume $1 + (n-1)\alpha \ne 0$, and $$ BJ = \frac{1}{1 + (n-1)\alpha} J $$ Plugging this back into (1), $$ \frac{\alpha}{1 + (n-1)\alpha} J + (1- \alpha) B = I $$ Now solve for $B$.

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