A I can't verify a relationship between cofactor and determinant

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In "General Relativity" by Hobson, et. al., in p. 66, we find a sentence starting "If we denote the value of the determinant ... " in the lower part of the page. I can not verify this assertion. Please help me the verification.
On that sentence, cofactor of an element of a metric is derived. But I can not verify it. Here I attached the copy of the page.
 

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stevendaryl

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It’s because the inverse of ##g_{\mu\nu}## is ##g^{\mu\nu}##, and the inverse of a matrix is given by the matrix of cofactors divided by the determinant.
 
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Rund and Lovelock ("Tensors, Differential Forms, and Variational Principles") have a good section on this in chapter 4

I hope you are familiar with Levi-Civita pseudo-tensors (https://en.wikipedia.org/wiki/Levi-Civita_symbol)

The determinant of the n-by-n matrix ##g_{\alpha\beta}## is given by:

##g = \epsilon^{\mu_1,\dots,\mu_n}g_{1\mu_1}\dots g_{n\mu_n}##

Now consider the following object:

##\omega^{\alpha\beta}=\sum_s\delta^\beta_s \epsilon^{\mu_1,\dots,\mu_{s-1},\alpha,\mu_{s+1},\dots,\mu_n}g_{1\mu_1}\dots g_{s-1,\mu_{s-1}}g_{s+1,\mu_{s+1}}\dots g_{n\mu_n}##

##g_{\kappa,\alpha}\omega^{\alpha\beta}=\sum_s \delta^\beta_s \epsilon^{\mu_1,\dots,\mu_n}g_{1\mu_1}\dots g_{s-1,\mu_{s-1}}\: g_{\kappa,\mu_s} \:g_{s+1,\mu_{s+1}}\dots g_{n\mu_n}##

Due to anti-symmetry of Levi-Civita, the only non-zero term in the ##\sum_s## is the one with ##s=\kappa##, but thet term is only non-zero if ##s=\beta=\kappa##. Now if all of this works, then we get the earlier expression for the determinant, so:

##g_{\kappa,\alpha}\omega^{\alpha\beta}=g\delta^\beta_\kappa##

So ##\omega^{\alpha\beta}=g g^{\alpha\beta}##, the inverse times the determinant. The co-factor matrix is here somewhere, but you don't need it to proceed. Note that from the above definition of the determinant

##\partial_c g = \partial_c g_{\beta \alpha}\sum_s\delta^\beta_s \epsilon^{\mu_1,\dots,\mu_{s-1},\alpha,\mu_{s+1},\dots,\mu_n}g_{1\mu_1}\dots g_{s-1,\mu_{s-1}}g_{s+1,\mu_{s+1}}\dots g_{n\mu_n} = g g^{\beta\alpha} \partial_c g_{\alpha\beta}##

Which is what you were after (once we use ##g_{\alpha\beta}=g_{\beta\alpha}##)
 
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Thank you for all the replies there. I will study the precious information and suggestions given by all of you. :smile:
 

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