How is the Jacobian derived using this method?. My course notes say the following: $$\left[\begin{array}{l}{d x} \\ {d y}\end{array}\right]=\left[\begin{array}{ll}{x_{u}} & {x_{v}} \\ {y_{u}} & {y_{v}}\end{array}\right]\left[\begin{array}{l}{d u} \\ {d v}\end{array}\right]\\$$
$$d A=\left|\begin{array}{ll}{x_{u}} & {x_{v}} \\ {y_{u}} & {y_{v}}\end{array}\right| d u d v\\$$
Apparently the second line follows from the first, but expanding out the algebra of the two equations from the first line and multiplying both equations together (as $dA = dx dy$) I don't get the same as calculating the determinant in the second line and multiplying by $dudv$. So how does the second line follow from the first?
 A: View $$\begin{bmatrix}
dx \\
dy
\end{bmatrix},
\begin{bmatrix}
du \\
dv
\end{bmatrix}$$
as bases for vector spaces, call them $\vec{e_{xy}}$ and $\vec{e_{uv}}$. Naturally the area of the parallelogram formed by $\vec{e_{xy}}$ is the same as the area formed by the basis $J\vec{e_{uv}}$ where $J_{uv}$ is the Jacobian. The area of the latter is the area of the parallelogram formed by $\vec{e_{uv}}$, stretched by the determinant of the matrix it is being multiplied by, which is $J$. So
$$dA=\det{(J)} du dv$$
A: For this case, a geometric/vector interpretation would be the most helpful. Assume that $dx$ and $dy$ are perpendicular vectors. Then the area between the two vectors is $dA=|dx\times dy|$, where $\times$ is the vector (cross) product. Similarly, $du$ and $dv$ are perpendicular vectors. The area obtained by the cross product is not necessarily $dA$, but a scaled version of it. The scaling factor is the Jacobian determinant.
Using the first equation $$dx=x_udu+x_vdv\\dy=y_u du+y_v dv$$
Now calculating $dA$ using the cross product:
$$dA=|dx\times dy|=|(x_udu+x_vdv)\times(y_u du+y_v dv)|\\=|x_uy_u du\times du+x_vy_vdv\times dv+x_uy_vdu\times dv+x_vy_udv\times du|$$
Now we use some properties of the cross product:
$$a\times b=-(b\times a)\\a\times a=0\\|a\times b|=|a||b|\sin\theta$$
In the last equation $\theta$ is the angle between $a$ and $b$ vectors. So in our case, when $du$ and $dv$ are perpendicular $|du\times dv|=du\ dv$.
Then going back to $dA$ we have: $$dA=|x_uy_v-x_vy_u|du\ dv$$
The last absolute value is the Jacobian determinant.
