Question 3

Frequently, we encounter a need for two (different) linear transformations to transform a parallelogram into a rectangular region. When this happens, we need to make choices like: $u=-7 x-5 y$ and $v=-6 x+4 y$.

For this example, find the inverse transformations:
\[
\begin{array}{l}
x= \\
y=
\end{array}
\]
and use these to calculate the Jacobian:
\[
\frac{\partial(x, y)}{\partial(u, v)}=
\]
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Step 1 ：Given the system of equations: \(u = -7x - 5y\) and \(v = -6x + 4y\)

Step 2 ：Multiply the first equation by 4 and the second equation by 5 to get: \(4u = -28x - 20y\) and \(5v = -30x + 20y\)

Step 3 ：Add these two equations together to get: \(4u + 5v = -58x\)

Step 4 ：Solve for x to get: \(x = -\frac{4u + 5v}{58}\)

Step 5 ：Substitute x into the first equation to get: \(u = -7x - 5y\) which simplifies to \(u = \frac{28u + 35v}{58} - 5y\)

Step 6 ：Rearrange to get: \(58u = 28u + 35v - 290y\) which simplifies to \(30u = 35v - 290y\)

Step 7 ：Solve for y to get: \(y = \frac{35v - 30u}{290}\)

Step 8 ：The inverse transformations are: \(x = -\frac{4u + 5v}{58}\) and \(y = \frac{35v - 30u}{290}\)

Step 9 ：Calculate the partial derivatives of x and y with respect to u and v to get: \(\frac{\partial x}{\partial u} = -\frac{4}{58}\), \(\frac{\partial x}{\partial v} = -\frac{5}{58}\), \(\frac{\partial y}{\partial u} = -\frac{30}{290}\), and \(\frac{\partial y}{\partial v} = \frac{35}{290}\)

Step 10 ：The Jacobian is: \(\frac{\partial (x, y)}{\partial (u, v)} = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u} = -\frac{140}{16820} + \frac{150}{16820} = \frac{10}{16820} = \frac{1}{1682}\)

Step 11 ：\(\boxed{\text{So, the Jacobian of the transformation is } \frac{1}{1682}}\)