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Matrix determinants and differentiation

by tickle_monste
Tags: determinants, differentiation, matrix
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Sep17-09, 01:09 PM
P: 70
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.
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Sep17-09, 01:26 PM
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Hurkyl's Avatar
P: 16,091
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.
Sep17-09, 01:43 PM
HW Helper
P: 1,371
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.

Sep17-09, 02:42 PM
P: 70
Matrix determinants and differentiation

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?
Sep17-09, 03:48 PM
HW Helper
P: 1,371
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.
John Creighto
Sep17-09, 04:35 PM
P: 813
Quote Quote by Hurkyl View Post
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.
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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