The article


*Alesker, S. (2003). Quaternionic Monge-Ampere equations. The Journal of Geometric Analysis, 13(2), 205-238.*


has the following CLAIM:



> 
> **Claim.** *Let $A$ be an invertible hyperhermitian matrix of order $n$. For any $i$, $1\le i\le n$, $$\left(A^{-1}\right)\_{ii} = \frac{1}{\det A} \det M\_{ii}(A) .$$*
> 
> 
> 


$M\_{ii}$ denotes the minor of a quaternionic matrix $A$ obtained from $A$ by deleting the $i$-th row and the $i$-th column.


I also found how to get the inverse matrix in:


*Chen, L. (1991). Inverse matrix and properties of double determinant over quaternion field. Sci. China Ser. A, 34(5), 528-540.*


But I still don't know how to show the claim. Any help?



How to calculate $(A^{-1})_{ii}$ for an invertible hyperhermitian quaternionic matrix $A$?
Determinants are tricky for matrices of quaternions, but they are not as bad when the matrix is hermitian. In that case, one can expand in the usual way along any row and get the same result.


See Theorem 5.1 here: "Cramer's rule for quaternionic systems of linear equations "Kyrchei, II, Journal of Mathematical Sciences, 155(6), 839--858, (2008). This is the result you want, and the paper has a detailed proof.


