Let $V$ a vector space over $\mathbb{R}$. As far as I understand, a bivector is an antisymmetric second rank tensor, thus an element of $V\otimes V$.
Let $\{\underline{e_i}\}_{i=1}^n$ and $\{\underline{e'_i}\}_{i=1}^n$ be two basis sets for $V$ such that (please note that I'm using Einstein's convention): \begin{equation} \underline{e_i}=\lambda_i^j\underline{e_j'} \quad i=1,2...n \end{equation} The wedge product of two vectors of $V$ is a bivector so in terms of components it should transform like a second order contravariant tensor (an element of $V\otimes V$).
Let $\underline{a}=a^i\underline{e_i}, \underline{b}=b^i\underline{e_i}$ two vectors in $V$ and $\{e^i\}_{i=1}^n$, $\{e'^i\}_{i=1}^n$ be the dual bases corresponding to the basis above, so we have: \begin{equation} e^i=\mu_j^ie'^j \quad i=1,2...n \quad\text{with}\quad \lambda_a^b\mu_b^c=\delta_a^c \end{equation} Then, we can see that:
\begin{equation} (\underline{a}\wedge\underline{b})^{ij}=\underline{a}\wedge\underline{b}(e^i, e^j)=\underline{a}\wedge\underline{b}(\mu_a^ie'^a, \mu_b^je'^b)=\mu_a^i\mu_b^j\thinspace\underline{a}\wedge\underline{b}(e'^a,e'^b)=\mu_a^i\mu_b^j(\underline{a}\wedge\underline{b})'^{ab} \tag{1} \end{equation}
Being an antisymmetric second rank tensor, it has only $3$ independent components, so considering $V=\mathbb{R}^3$, each bivector corresponds to a vector of $V$. Introducing Levi-Civita simbol, we can define this vector by its components with respect to a basis:
\begin{equation} \underline{a}\times\underline{b}=\epsilon_{ijk}a^jb^k\underline{e_i} \tag{2} \end{equation}
This kind of object is also referred to as pseudovector or axial vector. How does $(2)$ when we change the basis and how can we we write an equation analogous to $(1)$?
In simple words I am asking if the cross product in equation $(2)$ "transforms with two matrices", like equation $(1)$.
Please, forgive my lack of rigor, I'm quite new to tensor calculus and I am a physics students as well.