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Let $g$ be the determinant of the metric tensor.

I want to derive the following equation $g_{,\nu}=gg^{\lambda \mu}g_{\lambda \mu,\nu}$. It is said that $gg^{\lambda \mu}$ is a cofactor, but I can't understand why. To begin with, I'm not familiar with how to express the determinant of the metric tensor i.e. $g$. I know that $g$, the determinant is the sum of the cofactors multiplied with corresponding matrix element.

Can I get some illuminations on this?

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    $\begingroup$ I don't have time to check myself, but usually this always come into play with identities like these: en.wikipedia.org/wiki/Jacobi%27s_formula $\endgroup$ – JamalS Oct 17 '19 at 0:23
  • $\begingroup$ I think the matrix identity $\ln{\det{M}}=\text{tr}\ln{M}$ is the key. $\endgroup$ – G. Smith Oct 17 '19 at 0:24
  • $\begingroup$ @G.Smith It's taking me some time to understand. Can you elaborate? $\endgroup$ – Nuri Oct 17 '19 at 13:28
  • $\begingroup$ @JamalS I have followed your link, and with that I've written in my own way below. $\endgroup$ – Nuri Oct 17 '19 at 13:29
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    $\begingroup$ @YeonwookJung Differentiating gives $(\det{M})^{-1}\partial\det{M}=\text{tr}M^{-1}\partial M$. Now let $M$ be the metric tensor, and write this using index notation. $\endgroup$ – G. Smith Oct 17 '19 at 16:37
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Determinant of the metric, or any matrix can be expressed through Levi-Civita (relative) tensors, or through generalized Kroenecker deltas

I have written about it here

https://www.physicsforums.com/threads/i-cant-verify-a-relationship-between-cofactor-and-determinant.970419/post-6165630

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Let $A$ be a $n\times n$ matrix.

Recall that $$A \text{adj}(A)=\det(A)\cdot1$$ where $1$ is the identity matrix, $\text{adj}(A)=C^T$ is the adjoint matrix, $C_{ij}=(-1)^{i+j}M_{ij}$ is the cofactor matrix, $M_{ij}$ is the minor of the $A_{ij}$. This is justified because $$\det{A}=\sum_j A_{ij}C_{ij}=\sum_j A_{ij}adj_{ji}=(A\text{adj}(A))_{ii}.$$

Now consider $$d(\det(A))=\sum_{i,j}\frac{\partial\det(A)}{\partial A_{ij}}dA_{ij}$$ where $$\frac{\partial\det(A)}{\partial A_{ij}}=\frac{\partial\sum_k A_{ik} \text{adj}_{ki}}{\partial A_{ij}}=\sum_k\frac{\partial A_{ik}\text{adj}_{ki}}{\partial A_{ij}}=\sum_k \frac{\partial A_{ik}}{\partial A_{ij}}\text{adj}_{ki}+A_{ik}\frac{\partial \text{adj}_{ki}}{\partial A_{ij}}=\sum_k \delta_{jk}\text{adj}_{ki}=\text{adj}_{ji}$$ since $\text{adj}_{ji}(A)=C_{ij}$ is independent of $A_{ij}$ as it is the determinant of the matrix A with $i^{th}$ row and $j^{th}$ column removed. Thus we have $$d(\det(A))=\sum_{i,j}\text{adj}_{ji} dA_{ij}=\sum_{i,j} \text{adj}_{ij} A_{ji} = tr(\text{adj(A)} dA)=tr(\det(A)A^{-1}dA).$$ Finally putting $(A)=(g)$, we conclude $$g_{,\nu}=gg^{\mu\lambda}g_{\mu\lambda ,\nu}.$$

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