Let's suppose we have a $3 \times 3$ matrix $A=(a_{ij})$ whose elements $a_{ij}$ are objects whose product is not commutative: $a_{ij}·a_{kl}\neq a_{kl}·a_{ij}$. Then, which would be the formula for the determinant $|A|$?
Asked
Active
Viewed 1,169 times
1 Answers
2
In general, for a square matrix of dimension $n$ with coefficients in a ring $\mathcal A$, the determinant is given by:
$$\det(A) = \sum_{\sigma \in \mathfrak S_n} \epsilon(\sigma) \prod_{i=1}^n a_{\sigma(i),i}$$
so you can apply that to the case $n=3$. Now take care that several properties of determinants may not be true anymore if the ring $\mathcal A$ is not commutative! You can have a look at quasideterminant.
mathcounterexamples.net
- 71,758
-
Ok, could you please develop a little more that expression? Not sure about the meaning of $\mathfrak S, \sigma$ and $\epsilon$... – Quaerendo May 08 '20 at 14:08
-
Have a look at determinant where the formula is applied for the case $n=3$. – mathcounterexamples.net May 08 '20 at 14:12
-
You meant this expression? $|A|=a_{11} a_{22} a_{33}+a_{12} a_{23} a_{31}+a_{13} a_{21} a_{32}-a_{13} a_{22} a_{31}-a_{12} a_{21} a_{33}-a_{11} a_{23} a_{32}$ – Quaerendo May 08 '20 at 14:23
-
Exactly that!!! – mathcounterexamples.net May 08 '20 at 14:35