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VinnyCee

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**Solve u = 2x - 3y & v = -x + y in terms of x & y, then find Jacobian.**

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__Here is the problem__Solve the system [tex]u = 2x\;-\;3y,\;\;v = -x\;+\;y[/tex] for x and y in terms of u and v. Then find the Jacobian [tex]\frac{\partial\left(x,\;y\right)}{\partial\left(u,\;v\right)}[/tex].

Find the image under the trasformation [tex]u = 2x\;-\;3y,\;\;v = -x\;+\;y[/tex] of the parallelogram [tex]R[/tex] in the xy-plane with boundries [tex]x = -3,\;\;x = 0,\;\;y = x\;and\;y = x\;+\;1[/tex]

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__Here is what I have__[tex]y = -u\;-\;2v\;and\;x = -u\;-\;3v[/tex] I am not sure of these answers.

Assuming those are correct, I get the Jacobian to be [tex]-2[/tex]. This is where I am stuck, how do I do the second part? I can get the [tex]v-values[/tex] to be [tex]0\;and\;1[/tex] here,

[tex]y = x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;y = x\;+\;1[/tex]

[tex]-2v\;-\;u = -3v\;-\;u\;\;\;\;\;-2v\;-\;u = -3v\;-\;u\;+\;1[/tex]

[tex]v = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;v = 1[/tex]

But when I try and do the last part for the u values, I get [tex]u = u\;+\;3[/tex], which is impossible becaus then [tex]0 = 3[/tex].

Please help!

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