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Exercise 2.6.26

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${(K^{- 1})}_{ij} = j (n - i + 1) ∕ (n + 1)$ for $i \geq j$ (and symmetric): Multiply $K^{- 1}$ by $n + 1$ (the determinant of $K$ ) to see all whole numbers.

mathchakra

2021-12-20 21:15

Exercise 2.6.26

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