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Suppose I have a matrix $A$ as following:

$$A=\begin{bmatrix}1&a_{12}&\cdots&a_{1n}\\ a_{21}&1&\cdots&a_{2n}\\\vdots&\vdots&\ddots&\vdots\\a_{n1}&a_{n2}&\cdots&1\end{bmatrix}$$

Assuming $$\max_{1\le{i}\le{n},1\le{j}\le{n}}\left|a_{ij}\right|\ll1$$

How to obtain the inverse of $A$ in an explicit form?

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Use $$(I-A)^{-1} = I + A + A^2 + A^3 + \cdots$$ for $\|A\|<1$, where $\|A\|=\sup_{\|x\|=1} \|Ax\|$.

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    $\begingroup$ Actually, $A$ in your response is $I-A$ from the question :-) The condition $\|A\|_M=\max_{i,j}|a_{ij}|<1$ is ok for the series to converge because $\|A\|_M$ is a matrix norm. $\endgroup$ – Algebraic Pavel Dec 30 '13 at 15:52
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The answer provided by nayrb is the best you can expect.

However, if you do not want an infinite series, but a finite (but self-referential) form then the following may be useful $$ (I-A)^{-1} = I + A + A^2 + A^3 \cdots A^{n-1} + A^n (I-A)^{-1} $$

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