Bourbaki

Exercise 14, § 8

Tag 065F, permanent. Cite it as [Bourbaki, Tag 065F].

Algebra, III, 641

Show the identity

∣0111…110a1+a2a1+a3…a1+an1a2+a10a2+a3…a2+an1a3+a1a3+a20…a3+an………………1an+a1an+a2an+a3…0∣=(−1)n2n−1∑i=1na1a2…ai−1ai+1…an \begin{vmatrix} 0 & 1 & 1 & 1 & \ldots & 1 \\ 1 & 0 & a_1 + a_2 & a_1 + a_3 & \ldots & a_1 + a_n \\ 1 & a_2 + a_1 & 0 & a_2 + a_3 & \ldots & a_2 + a_n \\ 1 & a_3 + a_1 & a_3 + a_2 & 0 & \ldots & a_3 + a_n \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 1 & a_n + a_1 & a_n + a_2 & a_n + a_3 & \ldots & 0 \end{vmatrix} = (-1)^n 2^{n-1} \sum_{i=1}^n a_1 a_2 \ldots a_{i-1} a_{i+1} \ldots a_n

(use Exercise 13).

Solution

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Cites

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