# 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 $A \to J((JA)J) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\bigg(\bigg(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}\bigg)\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\bigg) =\begin{bmatrix} b & a \\ d & c\end{bmatrix} Nope just ends up at -1 equivalence$

# Slater Determinants

If we have the matrix of individual wavefunctions for a two electron system $\begin{bmatrix} \phi_1(r_1) & \phi_2(r_1) \\ \phi_1(r_2) & \phi_2(r_2) \end{bmatrix}$

We know that the antisymmetric wavefunction of that system is $\Psi(r_1,r_2)= \begin{vmatrix} \phi_1(r_1) & \phi_2(r_1) \\ \phi_1(r_2) & \phi_2(r_2) \end{vmatrix}$

But from equation $$4$$ we know that $$|A|A^{-1}=M$$, where $$M$$ is some matrix. However, each element of this matrix is also a determinant.

For growing numbers of electrons these states can be expressed in terms of the old states. $\Psi(r_1)=\phi_1(r_1) \\ \sqrt{2}\Psi(r_1,r_2)=\phi_1(r_1)\phi_2(r_2)-\phi_2(r_1)\phi_1(r_2) = \Psi(r_1)\phi_2(r_2)-\phi_2(r_1)\Psi(r_2) \\ \sqrt{6}\Psi(r_1,r_2,r_3)=\dots=\Psi(r_1,r_2)\phi_3(r_3)+\Psi(r_2,r_3)\phi_3(r_1)+\Psi(r_3,r_1)\phi_3(r_2)$

This ends with $\Psi(r_1\dots r_N)=\frac{1}{\sqrt{N!}}\sum_{i=1}^n \Psi(r_{i+1 mod N?},r_{i+2 mod N?}\dots r_{N mod N?})\phi_N(r_i)$

But we know $\Psi(r_1,\dots,r_N)=\prod_i \psi_{n_i}(r_i)$

# Combination

If we then have some matrix of individual particle states corresponding to $$\phi_{nm}=\phi_m(r_n)$$ $A=\begin{bmatrix} \phi_{11} & \phi_{12} & \phi_{13} \\ \phi_{21} & \phi_{22} & \phi_{23} \\ \phi_{31} & \phi_{32} & \phi_{33} \end{bmatrix} \\ \\ |A|=\Psi(r_1,r_2,r_3)=\frac{1}{\sqrt{6}}||\phi_1\phi_2\phi_3||$ we can calculate the inverse as \[A^{-1}=\frac{1}{|A|} \begin{bmatrix} \begin{vmatrix} \phi_{22} & \phi_{23} \\ \phi_{32} & \phi_{33}\end{vmatrix} & \begin{vmatrix} \phi_{13} & \phi_{12} \\ \phi_{33} & \phi_{32}\end{vmatrix} & \begin{vmatrix} \phi_{12} & \phi_{13} \\ \phi_{22} & \phi_{23}\end{vmatrix} \\ \begin{vmatrix} \phi_{23} & \phi_{21} \\ \phi_{33} & \phi_{31}\end{vmatrix} & \begin{vmatrix} \phi_{11} & \phi_{13} \\ \phi_{31} & \phi_{33}\end{vmatrix} & \begin{vmatrix} \phi_{13} & \phi_{11} \\ \phi_{23} & \phi_{21}\end{vmatrix} \\ \begin{vmatrix} \phi_{21} & \phi_{22} \\ \phi_{31} & \phi_{32}\end{vmatrix} & \begin{vmatrix} \phi_{12} & \phi_{11} \\ \phi_{32} & \phi_{31}\end{vmatrix} & \begin{vmatrix} \phi_{11} & \phi_{12} \\ \phi_{21} & \phi_{22}\end{vmatrix} \end{bmatrix} = \begin{bmatrix} \Psi(r_2,r_3) & \begin{vmatrix} \phi_{13} & \phi_{12} \\ \phi_{33} &