Skip to main content
6 events
when toggle format what by license comment
Dec 28, 2018 at 4:31 vote accept Question Machine
Dec 28, 2018 at 4:13 comment added LSpice @user44191, thanks; I knew that it was a good decision to leave it as a comment rather than an answer. (Maybe a better decision still would have been to wait to comment until after some computation.)
Dec 28, 2018 at 3:11 answer added user44191 timeline score: 27
Dec 28, 2018 at 2:37 comment added user44191 @LSpice I don't think it can; consider the matrix $\begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}$. Additionally, even on that set, the "constraint" isn't exactly "on $a$"; if the bottom-right 2 by 2 has determinant $0$, then $a$ doesn't affect the determinant.
Dec 28, 2018 at 2:20 comment added LSpice I think that it can just be cut out of $\mathbb A^6$ by thinking of it as $\begin{pmatrix} a & 0 & c \\ 0 & 1 & f \\ g & h & i \end{pmatrix}$, with $a$ constrained so that the determinant is 1.
Dec 28, 2018 at 0:54 history asked Question Machine CC BY-SA 4.0