Determinant of a 2nd rank tensor help and inverse! I have the following 3x3 matrix
$$U_{ij} = g_{ij} + \epsilon_{ijk}u_k$$
and I want to find its inverse using the fact that it can be written as the linear combination of its symmetric part and its antisymmetric part i.e.
$$ (U^{-1})^{ij} = Ag^{ij} + Bu^iu^j + C\epsilon_{ijk}u_k. $$ The obvious thing to do is to take
$U_{ij} (U^{-1})^{jk}=\delta_i^k$ but I cannot go further. Have I put the indices right in the last equation? How do I proceed? I can raise and lower stuff using $g$ by the way!

$$U_{ij} = g_{ij} + \epsilon_{ijk}u^k\tag{1}$$
$$ (U^{-1})^{jl} = Ag^{jl} + Bu^ju^l + C\epsilon^{jlm}u_m.\tag{2}$$
$$\delta_i^l=U_{ij} (U^{-1})^{jl} = g_{ij}Ag^{jl} + g_{ij}Bu^ju^l+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{3}$$
$$\delta_i^l= g_{ij}Ag^{jl} + B(g_{ik}u^k)(g^{lm}u_m)+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{4}$$
$$\delta_i^l= A g_{i}^l + u^ku_m(B g_{ik}g^{lm}- C\epsilon_{jik}\epsilon^{jlm})\tag{5}$$
Using 
$$\epsilon_{jik}\epsilon^{jlm}=\delta_i^l\delta_k^m-\delta_i^m\delta_k^l\tag{6}$$
We obtain:
$$\delta_i^l= A g_{i}^l + u^ku_m(B g_{ik}g^{lm}- C\delta_i^l\delta_k^m+C\delta_i^m\delta_k^l)$$
$$=A g_{i}^l +B u_iu^l- Cu^ku_k\delta_i^l+Cu^lu_i$$
$$=A g_{i}^l +(B+C) u_iu^l- Cu^ku_k\delta_i^l\tag{7}$$
Setting $i=l$ and $\delta_i^i=3, u^2=u^ku_k$, we have:
$$3=A g_{i}^i +(B-2C) u^2\tag{8}$$
