Any 2 - form of the form
\[dx_i \wedge dx_j\]
for any arguments \[(\mathbf{a} , \mathbf{b})=((a_1 , a_2, ,,,,a_n),(b_1,b_2,...,b_n)) \]
is antisymmetrical in those arguments, so that \[dx_i \wedge dx_j (\mathbf{a} , \mathbf{b}) = - dx_i \wedge dx_j (\mathbf{b} , \mathbf{a}) \]
Proof
\[\begin{equation} \begin{aligned} dx_i \wedge dx_j (\mathbf{a}, \mathbf{b}) &= det \left( \begin{array}{cc} dx_i(\mathbf{a}) & dx_i(\mathbf{b}) \\ dx_j(\mathbf{a}) & dx_j(\mathbf{b}) \end{array} \right) \\ &= det \left( \begin{array}{cc} a_i & b_i \\ a_j & b_j \end{array} \right) \\ &=a_i b_j -b_i a_j \end{aligned} \end{equation}\]
\[\begin{equation} \begin{aligned} dx_i \wedge dx_j (\mathbf{b}, \mathbf{a}) &= det \left( \begin{array}{cc} dx_i(\mathbf{b}) & dx_i(\mathbf{a}) \\ dx_j(\mathbf{b}) & dx_j(\mathbf{a}) \end{array} \right) \\ &= det \left( \begin{array}{cc} b_i & a_i \\ b_j & b_j \end{array} \right) \\ &=b_i a_j -b_j a_i \end{aligned} \end{equation}\]