Inverse and Exchange Matrix, Quantum Mechanics, Slater Determinants

Inverse

For a 3x3 non-singular matrix \(A\) with a determinant \(|A|\) defined by \[A=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}\] we can calculate the inverse as \[A^{-1}=\frac{1}{|A|} \begin{bmatrix} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix} & \begin{vmatrix} a_{13} & a_{12} \\ a_{33} & a_{32}\end{vmatrix} & \begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23}\end{vmatrix} \\ \begin{vmatrix} a_{23} & a_{21} \\ a_{33} & a_{31}\end{vmatrix} & \begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33}\end{vmatrix} & \begin{vmatrix} a_{13} & a_{11} \\ a_{23} & a_{21}\end{vmatrix} \\ \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix} & \begin{vmatrix} a_{12} & a_{11} \\ a_{32} & a_{31}\end{vmatrix} & \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix} \end{bmatrix}\]

If we define an operation that is \(RCR_{ij}(A):=R_{ij}\), remove the \(i^{th}\) column and the \(j^{th}\) row of \(A\), then this is expressible as \[A^{-1}=\begin{bmatrix} |R_{11}| & |R_{12}J| & |R_{13}| \\ |R_{21}J| & |R_{22}| & |R_{23}J| \\ |R_{31}| & |R_{32}J| & |R_{33}| \end{bmatrix}\]

Where \[J=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\]

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 \(|A|=-|AJ_2|=-|J_2A|=|J_2AJ_2|\). This property \(|A|=-|AJ_2|\) 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 \(J_1\equiv-1\). This makes some sense, as for any \(J_nJ_n=I\) and \(J_nJ_nA=A\).

We can further extrapolate to the inverse of a 1x1 matrix \(A=A_{11}\), taking the \(R_{11}\) 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 \(A_{11}\), which again is the inverse of the 1x1 matrix.

Proof \(J_1=-1\):

For any \(J_n\), \(n>1\), \(|J_n|=-1\) as \(J_n\) is defined to be an antidiagonal matrix.
\(|AB|=|A||B|\).
Therefore \(|AJ_n|=-|A|\).
If one extrapolates to the case \(n=1\), For the above to remain true, \(J_1=-1\)

Other Concept

Looking at the same formula we can define four 3x3 matrices, top left, top right, bottom left, bottom right \[A^{TL}=\begin{bmatrix} a_{22} & a_{13} & a_{12} \\ a_{23} & a_{11} & a_{13} \\ a_{21} & a_{12} & a_{11} \end{bmatrix} \\ A^{TR}=\begin{bmatrix} a_{23} & a_{12} & a_{13} \\ a_{21} & a_{13} & a_{11} \\ a_{22} & a_{11} & a_{12} \end{bmatrix} \\ A^{BL}=\begin{bmatrix} a_{32} & a_{33} & a_{22} \\ a_{33} & a_{31} & a_{23} \\ a_{31} & a_{32} & a_{21} \end{bmatrix} \\ A^{BR}=\begin{bmatrix} a_{33} & a_{32} & a_{23} \\ a_{31} & a_{33} & a_{21} \\ a_{32} & a_{31} & a_{22} \end{bmatrix}\]

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 \[A^{TL}=\begin{bmatrix} R_2^TP_2 & R_1^TP_1 & R_1^TP_2 \end{bmatrix} =A^{r(211)}_{p(212)}\\ A^{TR}=\begin{bmatrix} R_2^TP_1 & R_1^TP_2 & R_1^TP_1 \end{bmatrix} =A^{r(211)}_{p(121)}\\ A^{BL}=\begin{bmatrix} R_3^TP_2 & R_3^TP_1 & R_2^TP_2 \end{bmatrix} =A^{r(332)}_{p(212)}\\ A^{BR}=\begin{bmatrix} R_3^TP_1 & R_3^TP_2 & R_2^TP_1 \end{bmatrix} =A^{r(332)}_{p(121)}\]

These matricies are then such that \[\frac{1}{|A|}(A^{TL} \circ A^{BR} - A^{TR} \circ A^{BL}) = A^{-1}\]

Where \(\circ\) is the Hadamard product or element-wise product.

Imaginary Equivalence

If we have the parallel that \(AJ\) is to \(A\) what \(-a\) is to \(a\), then what of the parallell \(ia\) to \(a\), consider the operation \[A \to J((JA)J) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\bigg(\bigg(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}\bigg)\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\bigg) =\begin{bmatrix} b & a \\ d & c\end{bmatrix} Nope just ends up at $-1$ equivalence\]