Alternative Definition of the Quantum Determinant? Let $M_q(n)$ be the standard quantum matrices (over the complex numbers) with generators $u^i_j$ for $i,j = 1, \ldots ,N$, and reations 
$$
u^i_ju^k_j = qu^k_ju^i_j, \text{ for } i < k, ~~~~~~~ u^i_ju^k_j = q^{-1}u^k_ju^i_j, \text{ for } i > k, 
$$
and so on .... The determinant element $\mathrm{det}$ is defined by
$$
\mathrm{det} = \sum_{\pi \in S_n} (-q)^{\mathrm{l}(\pi)}u^1_{\pi(1)} \cdots u^N_{\pi(N)},
$$
where $S_N$ is the group of permutations on $N$ objects, and $\mathrm{l}(\pi)$ is the length of $\pi$. 
Can anyone find a slick way to show that one can also define $\mathrm{det}$ by 
$$
\mathrm{det} = \sum_{\pi \in S_n} (-q)^{-\mathrm{l}(\pi)}u^N_{\pi(N)} \cdots u^1_{\pi(1)}?
$$
I am almost sure it is true (it works for the lower orders) but I can't seem to be able to prove it.
 A: The good approach to prove anything about (q)-determinats was proposed by Y. I. Manin.
It is via (q)-Grassman algebra.
If I am understanding yours question correctly, then answer can be  obtained on the following route.
Consider (q)-Grassman variables $\psi_i \psi_j = -q \psi_j \psi_i , ~  i < j $ and $\psi_i^2=0$. Consider matrix $U$ with elements $u_{ij}$
Notation: consider variables  $\psi_i^U= \sum_k \psi_k U_{ki}$,
Standard Lemma (Manin): $det_q(M) \prod_i \psi_i = \prod_i \psi_i^U$ - this holds for any matrix  "U" - do not need not satisfy quantum group relations - no relation at all is necessary.
KEY OBSERVATION (Manin): If $U$ satisfy relations of quantum group, then
$\psi_i^U$ will q-commute again !!! i.e. $\psi_i^U \psi_j^U= -q \psi_j^U\psi_i^U, ~(\psi_i^U)^2=0 $. 
(Actually you need only "half" of the relation of quantum group for this lemma to be true.
We proposed to call such "half"-quantum matrices "q-Manin" matrices see http://arxiv.org/abs/0901.0235).
Now the question you are asking about become rather obvious.
Just consider the product $\psi_1^U\psi_2^U...\psi_n^U$  in the opposite 
order $=(-q)^{n(n-1)/2} \psi_n^U\psi_{n-1}^U...\psi_1^U$ and also pay attention
that variables in the opposite order become $q^{-1}$-commuting.
So treating all these power of $q$ correctly we should arrive to yours formula,
if I am not mistaking.
If you write me e-mail al. mysurname gmail dot com I can send you some some materials about
q-Manin matrices where we discuss things like that...
For q=1 - these q-Manin matrices are NOT commutative - but all theorems of linear algebra
can be extended to them in the form precisely like standard commutative.
See http://arxiv.org/abs/0901.0235
Algebraic properties of Manin matrices 1
A. Chervov, G. Falqui, V. Rubtsov 
A: If your Quantum (semi)group coacts on a finite dimensional Nichols algebra, then the top degree of the Nichols algebra is a 1-dimensional comodule, and so, it provides a group-like element that is the natural candidate for the Quantum determinant. Moreover, the non degenerated product of the Nichols algebra allows to prove an antípod fórmula (after inverting this grouplike). We develop these calculations when the bialgebra is given by the FRT construction in https://arxiv.org/abs/1805.11736
Also, some formulas for qdet are given there.
