Do lines between determinants pass through the inverse?

Question

Let A be a $2 \times 2$ matrix whose inverse also exists. If I was to draw a line from each of the 3 vertices (that are not the origin) of the determinant of A, to the 3 vertices of the determinant of A $^{-1}$, will the lines pass through the vertices of the identity matrix (1,0), (1,1) and (0,1)?

Answer 1

Writing $v_1=(1,0),v_2=(0,1),v_3=v_1+v_2$, you want $A^{-1}v,v,Av$ to be co-linear for each of $v=v_i$.

This is the same as having $v,Av,A^2v$ co-linear.

Letting $a=\mathrm{tr}\, A$ and $b=\det A$, then $A^2-aA +bI=0$. So that means $A^2v = aAv-bv$. This is on the same line as $Av$ and $v$ if and only if $a-b=1$. So this is only true when $\det A +1 = \mathrm{tr}\,A$.

It's a bit more complicated if you want to allow, say $Av_1,A^{-1}v_2,v_3$ to be colinear, but you'll still find counter-examples.