Calculating a determinant using Jacobi's second theorem

Asked May 16 '17 at 15:08

Active May 16 '17 at 18:53

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Prove using Jacobi's second theorem on determinants that

$$\begin{vmatrix} 0 & a & b & c \\ -a & 0 & d & e \\ -b & -d & 0 & f \\ -c & -e & -f & 0 \\ \end{vmatrix} = (af-be+cd)^2$$

I can easily prove it using Laplace expansion for determinants but have no idea how to prove it using Jacobi's second theorem. A corollary of the theorem tells that the determinant would be a perfect square of a polynomial of the elements. But nothing further. Any help will be appreciated.

For the theorem, have a look here: Jacobi's Second Theorem on Determinants

edited May 16 '17 at 18:53

ancient mathematician

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asked May 16 '17 at 15:08

Sntn

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As @Semiclassical said, this theorem is not known generally. Please give a reference. – Jean Marie May 16 '17 at 15:15
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The expression is the square of the so-called "Pffafian". You will find a very interesting (recent) answer in (https://artofproblemsolving.com/community/c7h540072_prove_that_determinant_is_a_perfect_square). – Jean Marie May 16 '17 at 15:20
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Have a look also at (https://math.stackexchange.com/q/165787). For an interesting link with graph matching, see (http://math.mit.edu/~goemans/18438S14/lec4-algmat.pdf). – Jean Marie May 16 '17 at 15:31
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Is this the theorem that you're referring to? Or is it this one? – Ben Grossmann May 16 '17 at 15:36
@Omnomnomnom It was the Wolfram link. Thanks. I've included it in my question. – Sntn May 16 '17 at 16:30

Calculating a determinant using Jacobi's second theorem

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