Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.