- #1
unfunf22
- 15
- 0
Homework Statement
Given n2 functions fij, each differentiable on an interval (a,b), define F(x) = det[fij] for each x in (a,b). Prove that the derivative F'(x) is the sum of the n determinants, F'(x) = [tex]\sum_{i=0}^n det(Ai(x))$.[/tex] where Ai(x) is the matrix obtained by differentiating the functions in the ith row of [fij(x)].
Homework Equations
To be honest I'm not completely sure what equations would be useful in this proof. I cannot get a good intuition on it.
I suppose Leibniz formula could be handy: http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants
Expansion by minors might be useful as well: http://en.wikipedia.org/wiki/Laplace_expansion
However, that is as far as my intuition will take me.
The Attempt at a Solution
This has been more of a sit and think process than a pen to paper one. I understand what I am supposed to do, but I have not the slightest idea what my intuition to this proof should be. I can't see how the determinant of the matrix would look (since I cannot diagonalize it or anything to make the determinant easy to calculate). I'm not very strong in my understanding of the Leibniz formula for determinants. I understand the idea behind minor's a bit better (delete i-row,j-column and take determinant of what is left). If I could get a feel for what the determinant of [fij(x)] would look like I might be able to see why the derivative would look like F'(x) = [tex]\sum_{i=0}^n det(Ai(x))$.[/tex]
I just need some intuition here, basically, and perhaps a good explanation of the weapons in my arsenal that would help me do this proof.
-Ian