Let $\gamma$ be an arclength parameterized Frenet curve. We define the Curvature function to be $$\kappa = \left\lVert\gamma''\right\rVert > 0$$ and the Torsion function to be $$ \tau = \left\langle T \times N,\, N'\right\rangle$$ Show that the adapted frame $F = \left(T, N, T \times N\right): I\to \mathbf{SO}\left(3,R\right)\,$ and calculate $$A = F^{-1}F^{'}$$ in terms of $\kappa$ and $\tau$.
So if anyone could help me with how exactly I show that the adapted frame belongs in the Rotation group that would be exceptional. I know some properties of the rotation group are that the transpose of an element ( a $3\times 3$ matrix) is equivalent to the inverse of that element and also that each element has determinant $1$. But am unsure how to show this with my frame. I imagine I consider the elements $\,\left(T, N, T \times N\right)\,$ to each be $3\times 1$ column vectors and thus I have a $3\times 3$ matrix but beyond that I'm a bit lost.
Also in calculating $A$ I'm a bit unsure how exactly to show that in terms of $\kappa$ and $\tau$, given that $\kappa$ is given in terms of the curve $\gamma$ while $\tau$ is given in terms of the Frenet frame I'm not sure what I should do to move forward. I get that I'll need to take the derivative of $F$ and also take its inverse (really its transpose) then multiply them but then that leaves me with the issue of how exactly to represent the elements of $T, N,$ and $T \times N$. $T$ is easy as $\gamma = \left(x, y, z\right)\,$ so then $T = \gamma' = \left(x', y', z'\right)\,$ but it's not so straightforward depicting the elements for $N$ and $T \times N$. What I tried to do was just choose arbitrary symbols for each vectors elements then remember one set of symbols is orthogonal to another is orthogonal to another but this is quite clumsy and doesn't elucidate much.