Matrix determinants and differentiation

  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. Hurkyl

    Hurkyl 15,987
    Staff Emeritus
    Science Advisor
    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. statdad

    statdad 1,478
    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. 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. statdad

    statdad 1,478
    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. 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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