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Matrix determinants and differentiation

  1. Sep 17, 2009 #1
    I'm having trouble understanding where this concept comes from:
    Step 1) If you start out with the following two equations

    v + log u = xy
    u + log v = x - y.

    Step 2) And then perform implicit differentiation, taking v and u to be dependent upon both x and y:
    (d will represent the partial derivative symbol):

    dv/dx + (1/u)du/dx = y
    du/dx + (1/v)dv/dx = 1

    I can do some simple Gaussian reduction and obtain:
    du/dx = [u(v-y)]/[uv-1] which is the same answer my book gives, the only difference is that my book uses this method to find du/dx:

    Step 3)
    det(Row 1:yu, u; Row 2: v, 1)/det(Row 1: 1, u; Row 2: v, 1)
    which reduces to: (yu)(1) - (u)(v))/(1)(1) - (u)(v).
    If the two matrices were A, and B, respectively, then:
    a11 = yu, a12 = u, a21 = v, a22 = 1.
    b11 = 1, b12 = u, b21 = v, b22 = 1
    (sorry, I don't know how to put a real matrix into here)

    And my problem is that I just don't understand where these matrices came from. I think this may be some formula from linear algebra that I just don't remember, but my book gives no reference to what it's doing, and goes directly from step 2 to step 3, so I'm really kind of lost right now. Any help would be appreciated.
  2. jcsd
  3. Sep 17, 2009 #2


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    Staff Emeritus
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    Gold Member

    It looks they used Cramer's rule to solve the linear system of equations, rather than Gaussian elimination. (They also cleared denominators before solving)

    I couldn't guess why -- Cramer's rule is rather good for proving theorems but rather bad for calculation. I suppose the 2x2 case isn't quite so bad, though.
  4. Sep 17, 2009 #3


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    Homework Helper

    Your system of equations for the derivatives is
    \frac 1 u \frac{du}{dx} + \frac{dv}{dx} & = y \\
    \frac{du}{dx} + \frac 1 v \frac{dv}{dx} & = 1

    Multiply the equations to clear fractions:

    \frac{du}{dx} + u \cdot \frac{dv}{dx} & = uy\\
    v \cdot \frac{du}{dx} + \frac{dv}{dx} & = v

    This can be put into matrix form as

    1 & u \\ v & 1
    {du}/{dx} \\ {dv}/{dx}
    \end{bmatrix} =
    uy \\ v

    and, as Hurkyl said, Cramer's rule was apparently used. My only reasoning about why it was used in this case: by using Cramer's rule the numerator and denominator of the solutions are easier to keep track of than they are when you use Gaussian elimination.
  5. Sep 17, 2009 #4
    I see, now it makes sense. Thanks for the help! Also, when I hover the mouse over those matrices you used it has some LaTeX script, can I just write that script into the text box on here? Or do you have do something else?
  6. Sep 17, 2009 #5


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    Homework Helper

    you need to enclose the latex markup in delimiters, like this:

    [ t e x ]
    your markup goes in here
    [/ t e x]

    I left the spaces in "tex" and "/tex" to avoid problems with my note. You do need the [] pair in each case.
  7. Sep 17, 2009 #6
    For two by two it doesn't matter but I thought Cramer's rule was good for symbolic calculations but not so good for numeric calculations.
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