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$|AB|=|A||B|$. \\  Therefore $|AJ_n|=-|A|$. \\  If one extrapolates to the case $n=1$, For the above to remain true, $J_1=-1$  \section{Other Concept}  Looking at the same formula we can define four 3x3 matrices, top left, top right, bottom left, bottom right \begin{equation}  A^{TL}=\begin{bmatrix}   a_{22} & a_{13} & a_{12} \\ a_{23} & a_{11} & a_{13} \\ a_{21} & a_{12} & a_{11}   \end{bmatrix} \\  A^{TR}=\begin{bmatrix}   a_{23} & a_{12} & a_{13} \\ a_{21} & a_{13} & a_{11} \\ a_{22} & a_{11} & a_{12}   \end{bmatrix} \\  A^{BL}=\begin{bmatrix}   a_{32} & a_{33} & a_{22} \\ a_{33} & a_{31} & a_{23} \\ a_{31} & a_{32} & a_{21}   \end{bmatrix} \\  A^{BR}=\begin{bmatrix}   a_{33} & a_{32} & a_{23} \\ a_{31} & a_{33} & a_{21} \\ a_{32} & a_{31} & a_{22}   \end{bmatrix}  \end{equation}  These matrices are constructed from the rows of the original matrix as such, if R_i is the ith row of the original matrix, and P_n is an operator which cycles that row forward n times we have \begin{equation}  A^{TL}=\begin{bmatrix} R_2^TP_2 & R_1^TP_1 & R_1^TP_2 \end{bmatrix} \\  A^{TR}=\begin{bmatrix} R_2^TP_1 & R_1^TP_2 & R_1^TP_1 \end{bmatrix} \\  A^{BL}=\begin{bmatrix} R_3^TP_2 & R_3^TP_1 & R_2^TP_2 \end{bmatrix} \\  A^{BR}=\begin{bmatrix} R_3^TP_1 & R_3^TP_2 & R_2^TP_1 \end{bmatrix}  \end{equation}  These matricies are then such that \begin{equation}  \frac{1}{|A|}(A^{TL} \circ A^{BR} - A^{TR} \circ A^{BL}) = A^{-1}  \end{equation}