Is there a neat way, apart from brute force expansion, to prove that the shoelace formula (excluding constant) $$\begin{array}|a\;&c\;&e\;&a\\b&d&f&b\end{array}$$ is equivalent to $$\begin{array}|a-c&b-d\\a-e\;\;&b-f\end{array}$$ ?
Note that the shoelace formula can be expanded as a sum of three determinants, i.e. $$\begin{array}|a&c\\b&d\end{array}\;+\;\begin{array}|c&e\\d&f\end{array} \;+\;\begin{array}|e&a\\f&b\end{array}$$
The Wikipedia article shoelace formula explains the meaning of the $2\times 4$ matrix. It refers to twice the signed area of the triangle with the three vertices $(a,b),(c,d),(e,f)$. However, if we translate the triangle so that the vertex $(a,b)$ moves to the origin, then the other two vertices move to $(c-a,d-b),(e-a,f-b)$. Now the determinant of the $2\times 2$ matrix they form is twice the area of the triangle they form with the origin. This is directly related to the magnitude of the cross product of two vectors being twice the area of the triangle formed by the two vectors and the origin.