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.



Alternative Definition of the Quantum Determinant?
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 


