Consider the following object $$M_{ij} = A_{il}B_{lj}-A_{jl}B_{li}$$ where repeated indicies imply summation and $A$ and $B$ are invertible anti-symmetric matrices. Now, clearly, $M$ is anti-symmetric (skew) in the index $i, j$. Now, I wish to isolate an instance of $A$ and for that I make use of inverses to write
$$A_{ik} = M_{ij}B^{-1}_{jk}-B^{-1}_{kj}A_{jl}B_{li}$$
Now, $A$ is an anti-symmetric matrix, however, the RHS of the above equation is not manifestly anti-symmetric and I doubt if it is even anti-symmetric. Can someone suggest a way to write $A$ in a manifestly anti-symmetric way or point out any errors in my manipulation?
Let me clarify by making a small change to the notation. Let $A_{ij} := A^i_j$ and let matrix multiplication take place by only bottom-top contraction i.e. $(AB)^i_j = A^i_kB^k_j$.
This means that we now have the following facts, $$M^i_j = A^i_kB^k_j - B^i_kA^k_j$$ and $$A^i_j = -A^j_i, \qquad B^i_j = -B^j_i$$
Now we can separate out $A^i_j$ in 2 different ways, \begin{align} A^i_kB^k_j &= M^i_j + B^i_kA^k_j\\ A^i_kB^k_j (B^{-1})^j_l &= M^i_j(B^{-1})^j_l + B^i_kA^k_j(B^{-1})^j_l\\ A^i_l &= M^i_j(B^{-1})^j_l + B^i_kA^k_j(B^{-1})^j_l \tag{1} \end{align} similarly we can have (from the $B^i_kA^k_j$ term) \begin{equation} A^i_l = (B^{-1})^i_jA^j_kB^k_l - (B^{-1})^i_jM^j_l \tag{2} \end{equation}
Let us test the anti-symmetry of eq. (1). Note that the inverse of an anti-symmetric matrix is also anti-symmetric. We have, \begin{align} A^i_l &= M^i_j(B^{-1})^j_l + B^i_kA^k_j(B^{-1})^j_l\\ &= (-M^j_i)(-(B^{-1})^l_j) + (-B^k_i)(-A^j_k)(-(B^{-1})^l_j)\\ &= (B^{-1})^l_jM^j_i - (B^{-1})^l_jA^j_kB^k_i\\ &= -(A^l_i) \tag{using eq.(2)} \end{align}
This meant that the anti-symmetry of eq. (1) was manifested with the help of eq. (2). But yes, on its own it doesn't seem obvious why it would be anti-symmetric. I hope this helps clarify your doubt.