asked in Philosophy SE

@NikeDattani Thank you for looking at the question! Yes I have checked for a large number of random matrices and it turns out that $\det(AB+A+I) <0$ and $\det(BA+B+I)<0$ is not possible

So yes my intuition is that it is not possible for both of them to be negative

However I have seen that othe rcases are possible that is

@NikeDattani Actually, the question seems to be deleted: mathoverflow.net/questions/428251/…

it is possible to have $\det(AB+A+I) >0$ and $\det(BA+B+1)>0$

It is possible to have $\det(AB+A+I)<0$ and $\det(BA+B+I)>0$

It is possible to have $\det(AB+A+I)>0$ and $\det(BA+B+I)<0$

under the assumptions that all the eigenvalues of $AB^2 $ and $A^2B$ are less than $1$ in absolute value

But I have intuition that $\det(AB+A+I)<0$ and $\det(BA+B+I) < 0$ is not possible under the assumption of all the eigenvalues of $AB^2, A^2B$ are less than one in absolute value.

No problem, this was was a spontaneous complaint/rant. While that I have other issues that I've flaged to moderators of this and other sites Stack Exchange, I have no special ussues for the question that I've edited on MathOverflow as user142929 and identifier ** 428251**. It's just the pity for another missed opportunity, for all us! Many thanks for you @NikeDattani and for the professor in comments.

@BAYMAX Okay I fixed the title.