Deduct formula for adjugate tensor I am trying to decuct the formula: 
$\text{adj}(\mathbf{T})_{ij} = 0.5 \epsilon_{ipq} \epsilon_{jrs} T_{pr} T_{qs}$
from the definition:
$\text{adj}(\mathbf{T})(\mathbf{u} \times \mathbf{v}) = (\mathbf{T} \mathbf{u}) \times (\mathbf{T} \mathbf{v})$
where $\mathbf{T}$ is an arbitray second order Tensor and $\mathbf{u},\mathbf{v}$ are arbitray vectors, $\times$ is the cross product and the dot product between a second order Tensor and a vector is denoted by $\mathbf{T} \mathbf{v}$.
I am stuck at this point:
$\text{adj}(\mathbf{T})_{om} u_i v_j \epsilon_{ijm} = T_{ij} T_{kl} \epsilon_{iko} u_j v_l$ 
and I don't know how to get rid of u and v to get the required formula?
Thanks for your Help.
 A: Here is my own answer.
Denoting the dot product between 2 Vectors by $\mathbf{a}\cdot\mathbf{b}$.
The cross product between 2 orthogonal basevectors can be written as follows:
$\mathbf{e}_r \times \mathbf{e}_s = \epsilon_{rsp} \mathbf{e}_p$
multiplying both sides by $\epsilon_{jsr}$ yields:
$ \epsilon_{jsr} \mathbf{e}_r \times \mathbf{e}_s = \epsilon_{jsr}\epsilon_{rsp} \mathbf{e}_p$
And since $\epsilon_{jsr}\epsilon_{rsp} = \delta_{js} \delta_{sp} - \delta_{jp}\delta_{ss} $
$\mathbf{e}_j = \frac{1}{2} \epsilon_{jrs} \mathbf{e}_r \times  \mathbf{e}_s$
Using this the ij compenent of $\text{adj}(\mathbf{T})$ can be deducted:
$\text{adj}(\mathbf{T})_{ij} = \mathbf{e}_i  \cdot (\text{adj}(\mathbf{T}) \mathbf{e}_i )=  \frac{1}{2} \epsilon_{jrs} \mathbf{e}_i \cdot ((\text{adj}(\mathbf{T})(\mathbf{e}_r \times  \mathbf{e}_s))$
The above given defintion holds for arbitary vectors and therefore also for orthogonal basevectors. Therefore:
$(\text{adj}(\mathbf{T})(\mathbf{e}_r \times  \mathbf{e}_s)) = (\mathbf{T} \mathbf{e}_r) \times (\mathbf{T}\mathbf{e}_s)$
$\text{adj}(\mathbf{T})_{ij} =  \frac{1}{2} \epsilon_{jrs} \mathbf{e}_i \cdot ((\mathbf{T} \mathbf{e}_r) \times (\mathbf{T}\mathbf{e}_s))$
Calculating the dot and cross products yields the wanted result:
$\text{adj}(\mathbf{T})_{ij}=\frac{1}{2}  \epsilon_{ipq} \epsilon_{jrs} T_{pr} T{qs}$
