The cross product of two vectors $\mathbf{x}, \mathbf{y} \in \mathbb{R}^3$ can be represented as a matrix product as follows, if $\mathbf{x} = (x_1, x_2, x_3)^{\top}$ then $\mathbf{x} \times \mathbf{y} = [\mathbf{x}]_{\times}\mathbf{y}$ where:
\begin{equation*}
[\mathbf{x}]_{\times} = \begin{pmatrix}
0 & -x_3 & x_2 \\
x_3& 0 & -x_1\\
-x_2&x_1 &0
\end{pmatrix}
\end{equation*}
Further more any skew-symmetric matrix has an associated bilinear form, If $M_{ij}$ is a skew symmetric matrix, then $B = \sum_{i, j} M_{ij} dx_i \otimes dx_j$ is a bilinear form satisfying $B(i, j) = -B(j, i)$.
As such I thought that the skew symmetry of $[\mathbf{x}]_{\times}$, results from a natural associated bilinear form. However, the cross product returns a vector, not a scalar, like a bilinear form should, at least as I have learnt. However, this is not a homework question, so there is no need to provide only hints.
Best Answer
For each fixed $x\in\mathbb{R}^3$ you get the skew symmetric matrix $[x]_\times$. This matrix does correspond to a bilinear form. It corresponds to the bilinear form $$H:\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R} \ , \ (a,b)\mapsto a^T[x]_\times b=a^T\cdot(x\times b)$$I think you may be wondering why the matrix $[x]_\times$ is skew-symmetric? By definition $x\times y=-(y\times x)$ and so this bilinear form $H$ is skew symmetric. That is, $H(a,b)=-H(b,a)$. Hence the matrix representation of $H$ is skew-symmetric.