Exercise 5, § 8

Algebra, III, 639

If the $a_i$ and $b_j$ are elements of a commutative field such that $a_i + b_j \neq 0$ for every ordered pair of indices $(i, j)$ , prove that

\det \left( \frac{1}{a_i + b_j} \right) = \frac{\sum_{i < j} (a_j - a_i)(b_j - b_i)}{\sum_{i,j} (a_i + b_j)}

("Cauchy's determinant").

Solution

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