$\|\mathbf{r}'\|$ in terms of Frenet-Serret Frame variables

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On Wikipedia I find the equation:

$$\frac{d}{dt} \begin{bmatrix} \mathbf{T}\\ \mathbf{N}\\ \mathbf{B} \end{bmatrix} = \|\mathbf{r}'(t)\| \begin{bmatrix} 0&\kappa&0\\ -\kappa&0&\tau\\ 0&-\tau&0 \end{bmatrix} \begin{bmatrix} \mathbf{T}\\ \mathbf{N}\\ \mathbf{B} \end{bmatrix}. $$

I was wandering if it was possible to replace $\|\mathbf{r}'(t)\|$ in terms of $\mathbf{T},\mathbf{N},\mathbf{B},\kappa,\tau$.

My attempt was to to expand the equation for $\frac{d}{dt}\mathbf{T}$ and take the absolute value on both sides. However, then we get $\|\mathbf{r}'(t)\|$ in terms of the absolute value of the derivative of the tangent which is not ideal in my case.

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No, that is not possible. All the other vectors and values are determined by the geometry of the curve, independent of the parametrization. The only place where the actual speed along the curve enters the equation is $\|{\bf r}'(t)\|$. This speed can not be reconstructed from the other terms.