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Carlo Beenakker
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PointersPerhaps the simplest approach first computes the determinant of an arbitrary Cauchy matrix, here is a derivation. Since every submatrix of a Cauchy matrix is also a Cauchy matrix, you can then find the inverse $H$ from the adjoint of $H$.

Some pointers to proofsworked-out calculations:

For a direct evaluation of the sum of all entries of $H^{-1}$, see The entry sum of the inverse Cauchy matrix

Pointers to proofs:

For a direct evaluation of the sum of all entries of $H^{-1}$, see The entry sum of the inverse Cauchy matrix

Perhaps the simplest approach first computes the determinant of an arbitrary Cauchy matrix, here is a derivation. Since every submatrix of a Cauchy matrix is also a Cauchy matrix, you can then find the inverse $H$ from the adjoint of $H$.

Some pointers to worked-out calculations:

For a direct evaluation of the sum of all entries of $H^{-1}$, see The entry sum of the inverse Cauchy matrix

Source Link
Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651

Pointers to proofs:

For a direct evaluation of the sum of all entries of $H^{-1}$, see The entry sum of the inverse Cauchy matrix