Special linear group of degree three or higher is not ambivalent

Statement
Suppose $$R$$ is a commutative unital ring. Then, the special linear group of degree three $$SL(3,R)$$ is not an disproves property satisfaction of::ambivalent group. In fact, for $$n \ge 3$$, the group $$SL(n,R)$$ is not an ambivalent group.

Case $$n = 3$$
Given a matrix $$A$$ with characteristic polynomial $$x^3 + kx^2 + lx -1$$, with $$m$$ invertible, the characteristic polynomial of $$A^{-1}$$ is $$x^3 - lx^2 - kx - 1$$. We need to choose values of $$k,l$$ such that these characteristic polynomials are distinct. Consider the case $$k = -1, l = 0$$. Thus, $$A$$ has characteristic polynomial $$x^3 - x^2 - 1$$ and $$A^{-1}$$ has characteristic polynomial $$x^3 + x -1$$.

Explicitly, we could choose:

$$A = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \\\end{pmatrix}, A^{-1} = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 0 & 0 \\\end{pmatrix}$$

These have different traces (1 and 0). Note that the example works even over field:F2 and rings of characteristic two.

Case of higher $$n$$
We can pad the example given above with the identity matrix, i.e., using $$A$$ and $$A^{-1}$$ as above, set:

$$B = \begin{pmatrix} A & 0 \\ 0 & I_{n-3} \\\end{pmatrix}, B^{-1} = \begin{pmatrix} A^{-1} & 0 \\ 0 & I_{n-3} \\\end{pmatrix}$$