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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?

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  • $\begingroup$ Maybe I'm missing something, but how is the claim any different from the standard Cramer's rule? Is the underlying field crucial here? $\endgroup$
    – Suvrit
    Oct 13, 2013 at 15:05
  • $\begingroup$ @suv....rit Quaternions do not commute, so I am not sure that everything works as we expect from linear algebra (in fact, since I know almost zilch on linear algebra on quaternions, I am not even sure how that determinant is defined). $\endgroup$ Oct 13, 2013 at 20:18
  • $\begingroup$ @Federico: Ah grazie! I was missing that part. Ok, then this is definitely an interesting question; also, I found the following: citeseerx.ist.psu.edu/viewdoc/… $\endgroup$
    – Suvrit
    Oct 13, 2013 at 21:09
  • $\begingroup$ This paper might also be useful: arxiv.org/pdf/math/0702447 $\endgroup$
    – Suvrit
    Oct 14, 2013 at 1:42

1 Answer 1

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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.

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