Consider Caley-Hamilton \[\det ( A) A^{-1} = \sum_{s=0}^{n-1}A^{s}\sum_{k_1,k_2,\ldots ,k_{n-1}}\prod_{l=1}^{n-1} \frac{(-1)^{k_l+1}}{l^{k_l}k_{l}!}\mathrm{tr}(A^l)^{k_l},\]

\[A^{-1} = \frac{1}{|A|}\sum_{s=0}^{n-1}A^{s}\sum_{k_1,k_2,\ldots ,k_{n-1}}\prod_{l=1}^{n-1} \frac{(-1)^{k_l+1}}{l^{k_l}k_{l}!}\mathrm{tr}(A^l)^{k_l},\]

Consider the use of an object \(|[A_{00},A_{11}]|=A_{00}A_{11}-A_{01}A_{10}\)... Much like a commutator, something like a detutator, represents the determinant of a 2x2 matrix when trace elementsw are inserted.

Then for a 3x3 matrix the similar thing \[|[A_{00},A_{11},A_{22}]| = |[A_{00},A_{11}]|A_{22}+|[A_{01},A_{12}]|A_{20}+|[A_{02},A_{10}]|A_{21}\]

The ordering of these being if the matrix were written next to itself, a diagonal shifting along each row.

We can write in summation form \[|[A_{00},A_{11},A_{22}]| = \sum_{i=0}^2 |[A_{0,0+i \;mod\; 3},A_{1,1+i \;mod\; 3}]|A_{2,2+i \;mod\; 3}\]

Therefore we can write the general formula \[|A|=|[A_{00},\dots,A_{nn}]| = \sum_{i=0}^{n} |[A_{0,0+i \;mod\; n+1},\dots,A_{n-1,n-1+i \;mod\; n+1}]|A_{n,n+i \;mod\; n+1}\]

Computational expense is bad 2x2, requires 2 multiplications and one subtraction, 3x3 requires 3 multiplications, 2 adds and 3(2x2’s), thus 9 multiplications 3 subtractions and 2 adds, 4x4 requires 4 multiplications, 3 adds and 4(3x3’s), thus 40 multiplications 12 subtractions and 11 adds.

For multiplications will be from\(n=0\), \(0,0,2,9,40,205,1236,...\) which is OEIS A038156 or \(O(n! \sum_{k=1}^{n-1} \frac{1}{k!} ))\) comparing this to \(n!\) as \(1,1,2,6,24,120\) is pretty bad. If one could only change the \(k=1\) to a \(k=2\) in the sum...

However, this construction would have it that \(\Psi(r_1,r_2)=|[\phi_1(r_1),\phi_2(r_2)]|\)