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Homework Help: Det(A) question

  1. May 2, 2007 #1
    1. The problem statement, all variables and given/known data
    I hope that I'm going to make sense here. I found the Jacobian of two functions by taking the partial of F1 w.r.t x, then y and same for F2.

    My professor did it by taking the partial of F1 w.r.t. y, then x, same for F2. So I have a 2x2 matrix. When i go to find the determinant, mine's negative and hers is positive. Isn't it supposed to be the same thing no matter which way you do it? this is driving me NUTS!

    It looks like this
    [-a -b]
    [-c d]
    so I get -ad-bc
    hers is
    [-b -a]
    [d -c]
    so she gets bc+ad
    What is the deal? What does it mean?
    2. Relevant equations

    w.r.t means with respect to
    3. The attempt at a solution
  2. jcsd
  3. May 2, 2007 #2

    Tom Mattson

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    No, certainly not. Simple example: It is a basic theorem of linear algebra that the determinant of a square matrix changes sign if any two rows (or columns) are interchanged. In a Jacobian determinant that would amount to swapping F1 and F2, while maintaining the order of x and y. You would certainly get a sign change in det(J) under that condition.

    You should not expect the determinant to remain invariant under permutation of the matrix elements, in general.
  4. May 2, 2007 #3


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    I think happyg1 knows that, but is concerned about the fact that the Jacobian is the conversion factor between area elements, e.g. [itex]dxdy=Jdrd\varphi[/itex]. He thinks it's weird that there are two possible answers to what J is.

    A partial answer is that you should think of area elements as "oriented", i.e. as if they have an "up" side and a "down" side. When the "up" side is up, the area is positive, when the "up" side is down, the area is negative. I have to go to bed now, so I don't have time to explain that further right now, but perhaps someone else will.
  5. May 2, 2007 #4
    I'm using this jacobian to determine steady states of a herbivore/plant model.
    When the trace is negative and the determinant is positive, we get a stable steady state. When the trace is negative (trace is negative in both cases) and the determinant is negative, we get an unstable steady state, so it appears to me that depending in the order that you do you partials dictates the stability of the model....but that makes no sense! If it's stable, it shouldn't matter which way you do the partials, the same result should fall out. In this case, it doesn't. I don't get it.

    Any thoughts on how to resolve this? I get unstable with my Jacobian and she gets stable with hers....I guess since biological models are dealing with positive things (like area, number of aphids and level of toxins) that the determinant should be positive in this case?

    Yes? No?

    Any input will be appreciated.

    EDIT I did partial of x first and she did partial of y. F1 and F2 weren't swapped. It's partial of F1 w.r.t x first vs. partial of F1 w.r.t. y first...but it's the same thing.. my columns are switched from hers.
    Last edited: May 2, 2007
  6. May 2, 2007 #5


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    If you are using the jacobian to compute, for example, the area of a map image then you don't care about sign (orientation). In which case you just take the absolute value of the jacobian anyway. So yes, if you are dealing with aphids, take the absolute value.
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