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I'm new to index notation and i'm trying to prove the identity $$\frac{d}{dt}\det(A(t))=\det(A)\operatorname{tr} \left(A^{-1}\frac{dA}{dt}\right) $$ for the special case when $A$ is an invertible second order tensor, using just index notation in the following fashion: \begin{align} \frac{d}{dt}\det(A(t)) & =\frac{d}{dt}(\varepsilon_{ijk}A_{i1}A_{j2}A_{k3}) \\[10pt] & =\varepsilon_{ijk} \left(\frac{d}{dt}(A_{i1})A_{j2}A_{k3}+\frac{d}{dt}(A_{j2})A_{i1}A_{k3}+\frac{d}{dt}(A_{k3})A_{i1}A_{j2})\right) \\[10pt] & =\varepsilon_{ijk}A_{r1}A_{j2}A_{k3} \left( A^{-1}_{1r}\frac{A_{i1}}{dt} + A^{-1}_{2r} \frac{A_{j2}}{dt}+A^{-1}_{3r}\frac{A_{k3}}{dt} \right) \end{align} where $\varepsilon$ is the permutation symbol. This is as far as I can get and it frustrates me because if I just switch the index "$r$" to "$i$" then the proof is done. Have I missed something obvious or made some error? I don't know how to proceed from here.

Best regards
Bengt

Michael Hardy
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Bengt
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    Hint: $\det(A)A^{-1}$ is the adjugate matrix of $A$. See e.g. https://en.wikipedia.org/wiki/Adjugate_matrix#Jacobi's_formula – Zhuoran He Feb 21 '18 at 18:05
  • I will use the theory you propose when proving this identity for a invertible tensor of order n. The present case is mostly so that I can practice using index notation, do you have any hints in that regard? – Bengt Feb 21 '18 at 18:13
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    The index notation looks like a dead end to me, because $(A_{ij})^{-1}\neq (A^{-1})_{ij}$. One has to find a way to introduce the inverse matrix $A^{-1}$ rather than inverse of the matrix entries. – Zhuoran He Feb 21 '18 at 18:21
  • When you write \text{det} instead of \det, then you see $3\text{det} A$ instead of $3\det A.$ The spacing is context-dependent, so that the amount of space to the right of $\det$ in $3\det A$ is larger than that in $3\det(A).$ \text{} does not have that feature. With $\operatorname{tr} A$ I've used \operatorname{} in my edits to your question. – Michael Hardy Feb 21 '18 at 18:24
  • @ZhuoranHe This is a good observation, but it turns out that there is a way to introduce the inverse matrix in index notation. – Oscar Cunningham Feb 22 '18 at 10:46

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You have to know the formula for the inverse matrix in index notation: $$\left(A^{-1}\right)_{1i}=\frac{\varepsilon_{ijk}A_{j2}A_{k3}}{\det(A)}$$ and similarly with $1$, $2$ and $3$ cycled. See also here.

So $$\varepsilon_{ijk} \left(\frac{d}{dt}(A_{i1})A_{j2}A_{k3}+\frac{d}{dt}(A_{j2})A_{i1}A_{k3}+\frac{d}{dt}(A_{k3})A_{i1}A_{j2})\right)$$ $$=\frac{d}{dt}(A_{i1})\left(A^{-1}\right)_{1i}\det(A)+\frac{d}{dt}(A_{j2})\left(A^{-1}\right)_{2j}\det(A)+\frac{d}{dt}(A_{k3})\left(A^{-1}\right)_{3k}\det(A)$$ $$=\det(A)\left(\left(A^{-1}\right)_{1i}\frac{d}{dt}(A_{i1})+\left(A^{-1}\right)_{2j}\frac{d}{dt}(A_{j2})+\left(A^{-1}\right)_{3k}\frac{d}{dt}(A_{k3})\right)$$ $$=\det(A)\left(\left(A^{-1}\right)_{li}\frac{d}{dt}(A_{il})\right)$$ $$=\det(A)\operatorname{tr}\left(A^{-1}\frac{dA}{dt}\right)$$

Oscar Cunningham
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