Matrix determinants and differentiation

1. tickle_monste

69
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. Hurkyl

15,998
Staff Emeritus
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.

1,478
t_m:
Your system of equations for the derivatives is
\begin{align*} \frac 1 u \frac{du}{dx} + \frac{dv}{dx} & = y \\ \frac{du}{dx} + \frac 1 v \frac{dv}{dx} & = 1 \end{align*}

Multiply the equations to clear fractions:

\begin{align*} \frac{du}{dx} + u \cdot \frac{dv}{dx} & = uy\\ v \cdot \frac{du}{dx} + \frac{dv}{dx} & = v \end{align*}

This can be put into matrix form as

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

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.

4. tickle_monste

69
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?

1,478
you need to enclose the latex markup in delimiters, like this:

[ t e x ]