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I was attempting to solve: $$ -2dx_{1} \land dx_{4}\left( \begin{bmatrix} 2 \\ 3 \\ -5 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ 4 \\ -5 \end{bmatrix}\right) $$ I solved this by pulling out the scalar and then finding the determinant of the matrix given by (let $u$ be the first column vector and $v$ be the following one): $$ \det\begin{bmatrix} -2*dx_1(u) & -2*dx_1(v) \\ dx_4(u) & dx_4(v) \end{bmatrix}=-2*\det\begin{bmatrix} dx_1(u) & dx_1(v) \\ dx_4(u) & dx_4(v) \end{bmatrix}=-2*\det\begin{bmatrix} 2 & 2 \\ 2 & -5\end{bmatrix}=-2(-14)=28$$ We then have to explain what this number represents and that is where my confusion comes in. It is my understanding that this represents the area of the parallelogram formed from $u$ and $v$ being projected onto the $\partial_{x_1}\partial_{x_4}$ plane. Should this area be positive or negative tho? Without the scalar $-2$ the oriented area would flip since going from $u$ to $v$ would be clockwise but with the scalar it would be counterclockwise so you wouldn't switch the sign? I guess my main confusion is coming from whether or not the determinant has the orientation built in and whether or not the scalar on one of the one-forms affects the orientation?

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  • $\begingroup$ What happens if you swap the two vectors? $\endgroup$ – amd Feb 27 at 18:44

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