INVERSE For a 3x3 non-singular matrix A with a determinant |A| defined by A= a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} we can calculate the inverse as A^{-1}={|A|} a_{22} & a_{23} \\ a_{32} & a_{33} & a_{13} & a_{12} \\ a_{33} & a_{32} & a_{12} & a_{13} \\ a_{22} & a_{23} \\ a_{23} & a_{21} \\ a_{33} & a_{31} & a_{11} & a_{13} \\ a_{31} & a_{33} & a_{13} & a_{11} \\ a_{23} & a_{21} \\ a_{21} & a_{22} \\ a_{31} & a_{32} & a_{12} & a_{11} \\ a_{32} & a_{31} & a_{11} & a_{12} \\ a_{21} & a_{22} If we define an operation that is RCRij(A):=Rij, remove the ith column and the jth row of A, then this is expressible as A^{-1}= |R_{11}| & |R_{12}J| & |R_{13}| \\ |R_{21}J| & |R_{22}| & |R_{23}J| \\ |R_{31}| & |R_{32}J| & |R_{33}| Where J= 0 & 1 \\ 1 & 0 Clearly the condition for a right multiplication of the J matrix being i + j = odd. Alternatively |A|(A^{-1})_{ij}=|R_{ij}J^{(i+j-1)}|=|J^{(i+j-1)R_{ij}}| This likely works because for any 2x2 matrix |A|= − |AJ₂|= − |J₂A|=|J₂AJ₂|. This property |A|= − |AJ₂| also appears to hold for a 3x3 matrix. Extrapolating backwards for a two by two matrix we get the correct formula on the proviso we define J₁ ≡ −1. This makes some sense, as for any JnJn = I and JnJnA = A. We can further extrapolate to the inverse of a 1x1 matrix A = A₁₁, taking the R₁₁ element to be the zero matrix, the determinant of this matrix is 1 and the reciprocal of the determinant of A is then just the reciprocal of A₁₁, which again is the inverse of the 1x1 matrix. Proof J₁ = −1: For any Jn, n > 1, |Jn|= − 1 as Jn is defined to be an antidiagonal matrix. |AB|=|A||B|. Therefore |AJn|= − |A|. If one extrapolates to the case n = 1, For the above to remain true, J₁ = −1 OTHER CONCEPT Looking at the same formula we can define four 3x3 matrices, top left, top right, bottom left, bottom right A^{TL}= a_{22} & a_{13} & a_{12} \\ a_{23} & a_{11} & a_{13} \\ a_{21} & a_{12} & a_{11} \\ A^{TR}= a_{23} & a_{12} & a_{13} \\ a_{21} & a_{13} & a_{11} \\ a_{22} & a_{11} & a_{12} \\ A^{BL}= a_{32} & a_{33} & a_{22} \\ a_{33} & a_{31} & a_{23} \\ a_{31} & a_{32} & a_{21} \\ A^{BR}= a_{33} & a_{32} & a_{23} \\ a_{31} & a_{33} & a_{21} \\ a_{32} & a_{31} & a_{22} 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 A^{TL}= R_2^TP_2 & R_1^TP_1 & R_1^TP_2 =A^{r(211)}_{p(212)}\\ A^{TR}= R_2^TP_1 & R_1^TP_2 & R_1^TP_1 =A^{r(211)}_{p(121)}\\ A^{BL}= R_3^TP_2 & R_3^TP_1 & R_2^TP_2 =A^{r(332)}_{p(212)}\\ A^{BR}= R_3^TP_1 & R_3^TP_2 & R_2^TP_1 =A^{r(332)}_{p(121)} These matricies are then such that {|A|}(A^{TL} \circ A^{BR} - A^{TR} \circ A^{BL}) = A^{-1} Where ∘ is the Hadamard product or element-wise product.