Register to reply 
Matrix determinants and differentiation 
Share this thread: 
#1
Sep1709, 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(vy)]/[uv1] 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
Sep1709, 01:26 PM

Emeritus
Sci Advisor
PF Gold
P: 16,092

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. 


#3
Sep1709, 01:43 PM

HW Helper
P: 1,362

t_m:
Your system of equations for the derivatives is [tex] \begin{align*} \frac 1 u \frac{du}{dx} + \frac{dv}{dx} & = y \\ \frac{du}{dx} + \frac 1 v \frac{dv}{dx} & = 1 \end{align*} [/tex] Multiply the equations to clear fractions: [tex] \begin{align*} \frac{du}{dx} + u \cdot \frac{dv}{dx} & = uy\\ v \cdot \frac{du}{dx} + \frac{dv}{dx} & = v \end{align*} [/tex] This can be put into matrix form as [tex] \begin{bmatrix} 1 & u \\ v & 1 \end{bmatrix} \, \begin{bmatrix} {du}/{dx} \\ {dv}/{dx} \end{bmatrix} = \begin{bmatrix} uy \\ v \end{bmatrix} [/tex] 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
Sep1709, 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?



#5
Sep1709, 03:48 PM

HW Helper
P: 1,362

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. 


#6
Sep1709, 04:35 PM

P: 813




Register to reply 
Related Discussions  
Matrix determinants and inverses  Calculus & Beyond Homework  8  
Easy matrix/determinants question  Calculus & Beyond Homework  1  
Trace of a matrix equals sum of its determinants?  Linear & Abstract Algebra  2  
Determinants of a matrix  Precalculus Mathematics Homework  7  
Determinants 4x4 matrix  Precalculus Mathematics Homework  2 