### WEDGE or EXTERIOR PRODUCT:

In reference to the explanation of the exterior or wedge product by Prof. Shifrin’s series (Chapter 8 - here, here and here), and connecting it with the more general tensor algebra operations, here is a summary:

The dual vector space $$V^*=(\mathbb R^n)^*$$ is a vector space of linear maps $$\mathbb R^n \to \mathbb R$$ that can be represented by a $$1\times n$$ matrix (i.e. row vector). It is isomorphic to $$\mathbb R^n,$$ and forms a vector space, which basis can be expressed as $$\mathrm e^1,\mathrm e^2,\dots,\mathrm e^n,$$ or as $$\mathrm dx_1, \mathrm dx_2, \dots ,\mathrm dx_n,$$ such that $$\mathrm dx_i(\vec v)=\vec {\mathrm e}_i\cdot \vec v.$$

The wedge product of two such forms in $$\Lambda^1(\mathbb R^3)^*$$ will result in an element in $$\Lambda^2(\mathbb R^3)^*.$$ For example, taking the last element of $$\Lambda^1(\mathbb R^3)^*$$ in the preceding example, i.e. $$\psi =4\mathrm dx_1 +7 \mathrm dx_2,$$ and wedging it with $$\varphi =2\mathrm dx_1 +5 \mathrm dx_3,$$ also in the same vector space $$\Lambda^1(\mathbb R^3)^*,$$ we obtain

$\psi\wedge\varphi=4\mathrm dx_1 +7 \mathrm dx_2 \wedge 2\mathrm dx_1 +5 \mathrm dx_3=-14\;\mathrm dx_{12}+20\;\mathrm d_{13}+35\; \mathbb d_{23}\tag 1$

Critically, this is the same as picturing the two forms in $$\Lambda^1(\mathbb R^3)^*,$$ i.e. $$4\mathrm dx_1 +7 \mathrm dx_2=\color{red}{\begin{bmatrix}4&7&0\end{bmatrix}}^\top$$ and $$2\mathrm dx_1 +5 \mathrm dx_3=\color{blue}{\begin{bmatrix}2&0&5\end{bmatrix}}^\top$$ as

$\begin{bmatrix} \color{red}4&\color{red}7&\color{red}0\\ \color{blue}2& \color{blue} 0 & \color{blue}5 \end{bmatrix}$

and realizing that $$\mathrm d_{12}$$ is simply the determinant of the first two columns (minor of the submatrix):

$\det{\begin{bmatrix} \color{red}4&\color{red}7\\ \color{blue}2& \color{blue} 0 \end{bmatrix}}=-14$

and that $$\mathrm d_{13}$$

$\det{\begin{bmatrix} \color{red}4&\color{red}0\\ \color{blue}2& \color{blue} 5 \end{bmatrix}}=20$

with $$\mathrm d_{23}$$

$\det{\begin{bmatrix} \color{red}7&\color{red}0\\ \color{blue}0& \color{blue} 5 \end{bmatrix}}=35$

We could express this as

$\psi\wedge\varphi=\sum \text {minors}_{ij}\; \mathrm d_{ij}$

The wedge product can be carried out between vector spaces $$\Lambda^l(\mathbb R^n)^*\wedge \Lambda^k(\mathbb R^n)^*\in \Lambda^{l+k}(\mathbb R^n)^*.$$

This all comes together when two vectors in $$\mathbb R^3$$ are fed into the expression in Eq (1), for example $\small\vec p =\begin{bmatrix}-1&-\pi&-\sqrt 2\end{bmatrix}^\top$ and $\small\vec q =\begin{bmatrix}10^1&10^2&10^3\end{bmatrix}^\top:$

\begin{align}\left(\psi \wedge \varphi\right)(p,q) &= -14\,\mathrm dx_{12}(p,q) + 20 \,\mathrm dx_{13}(p,q) + 35\, \mathrm dx_{23}(p,q)\\[2ex] &=-14 \,\begin{vmatrix}-1&10 \\-\pi & 10^2\end{vmatrix}+ 20\, \begin{vmatrix}-1 & 10 \\-\sqrt 2 & 10^3\end{vmatrix} + 35\, \begin{vmatrix}-\pi & 10^2 \\-\sqrt 2 & 10^3\end{vmatrix} \end{align}

As for the equivalence to tensor algebra pointed out in the answer

\begin{align} \mathrm dx_{12}(p,q)&=\mathrm dx_1 \wedge \mathrm dx_2 (p,q)\\[2ex] &= \left(\mathrm dx_1 \otimes \mathrm dx_2 - \mathrm dx_2\otimes \mathrm dx_1\right)(p,q)\\[2ex] &= (\mathrm dx_1 \otimes \mathrm dx_2 )(p,q)-(\mathrm dx_2\otimes \mathrm dx_1)(p,q)\\[2ex] &=\mathrm dx_1(p)\,\mathrm dx_2(q) - \mathrm dx_2(p) \,\mathrm dx_1(q)\\[2ex] &= (-1)(10^2) - (-\pi)(10)\\[2ex] &=\begin{vmatrix}-1&10 \\-\pi & 10^2\end{vmatrix}. \end{align}