Determinant of a 2nd rank tensor help and inverse!

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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!

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$$U_{ij} = g_{ij} + \epsilon_{ijk}u^k\tag{1}$$

$$ (U^{-1})^{jl} = Ag^{jl} + Bu^ju^l + C\epsilon^{jlm}u_m.\tag{2}$$

$$\delta_i^l=U_{ij} (U^{-1})^{jl} = g_{ij}Ag^{jl} + g_{ij}Bu^ju^l+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{3}$$

$$\delta_i^l= g_{ij}Ag^{jl} + B(g_{ik}u^k)(g^{lm}u_m)+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{4}$$

$$\delta_i^l= A g_{i}^l + u^ku_m(B g_{ik}g^{lm}- C\epsilon_{jik}\epsilon^{jlm})\tag{5}$$

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: $$3=A g_{i}^i +(B-2C) u^2\tag{8}$$