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Let $\mu_1, \ldots, \mu_n$ be unknowns. Let $C=(c_{ij})_{n \times n}$ be an invertible matrix. Suppose that $\sum_{\beta=1}^{n} \mu_{\beta} c_{\alpha, \beta}=\pi i$. I think that we can solve this equation by multiply $C^{-1}$ on both sides of the equation. It is said that the solution is $\mu_{\beta} = i \pi \sum_{\alpha=1}^n c_{\beta, \alpha}^{-1}$. But the solution I get is much more complicated.

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Perhaps there is a confusion between $[C^{-1}]_{ij}$ and $[C_{ij}^{-1}]$. –  anon May 17 '12 at 17:11
    
@anon, yes, you are right. –  LJR May 17 '12 at 17:16
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1 Answer

up vote 0 down vote accepted

With $\mu_b:=i\pi\sum_{a=1}^nc_{b,a}^{—1}$, we have $$\sum_{b=1}^n\mu_bc_{a,b}=\sum_{b=1}^n\sum_{a=1}^ni\pi c_{b,a}^{—1}c_{a,b}=\sum_{b=1}^ni\pi (CC^{—1})_{bb}=ni\pi,$$ so we actually have to take $\mu_b:=\frac 1ni\pi\sum_{a=1}^nc_{b,a}^{—1}$.

The fact you get a more complicated expression could be explained by the expression of the entries of the inverse matrix.

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