I have this 3x3 matrix $$E_{ij} = g_{ij} + \bar{\epsilon}_{ijk}z^k$$ and want to derive its inverse. I know that its inverse is given by $$(E^{-1})^{ij} = \frac{1}{1+z^2}(g^{ij} + z^{ij} - \bar{\epsilon}^{ijk}z_k) $$ (and yes, here the k index seems to be following Einstein convention-or what? ).
i) One way to find this is to say that the inverse will be a superposition of the symmetric part of $E$ and its anti-symmetric part. So it could be $$(E^{-1})^{ij} = Ag^{ij} + Bz^iz^j + C\bar{\epsilon}^{ijk}z_k$$ and then take the product $E_{ij}(E^{-1})^{jl} = \delta_{i}^{l}$. I am trying to do this but I cannot get enough equations to determine $A,B,C$. Can someone help with this?
ii) If I just try to take the first two equations and put them together to make sure they give me the unit matrix I arise to $1+z^2 + \bar{\epsilon}_{ljk}\bar{\epsilon}^{jlm}z^kz_m = 1+z^2$ which gives $$1+z^2 + \delta_{k}^{m}z^kz_m = 1+z^2$$ which is not right. The only other thing worth of mentioning is that $\bar{\epsilon}_{ijk} = \sqrt{\text{det}g} \epsilon_{ijk}$.
I basically care mostly about i). Can someone help me with this calculation?
Update: I think that the inverse given contains a mistake.
$E_{ij} (E^{-1})^{jk} = (A-Cz^2) \delta_i^k + (B+C)z_iz^k +(A+C) \varepsilon _{ijm}g^{jk} z_m$.
According to the definition, $E_{ij} (E^{-1})_{jk} = \delta_i^k$, so we must have : $(B+C) = (A+C)=0, (A-Cz^2)=1$, that is the result.
we use $ \varepsilon _{ijk} \varepsilon ^{ilm} = (\delta^l_j \delta^m_k - \delta^l_k \delta^m_j )$ in the computation.