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