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(A^{-1})_{ij}=|R_{ij}J^{(i+j-1)}|  \end{equation} Extrapolating backwards for a two by two matrix we get the correct formula on the proviso we definfe $J_1\equiv-1$. This makes some sense, as for any $J_nJ_n=I$ and $J_nJ_nA=A$.