- #1
Tomer
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Homework Statement
Hello, thanks for reading.
So, here's the problem. I'm given n2 functions fij(x), and the matrix A = (fij)
I get the definition of a new function: F(x) = detA
I need to prove, that F'(x) = [itex]\sum^{n}_{k=1}det(A_{k})[/itex]
Where Ak is a matrix identical to A, with only the k'th row being replaces with the row of the derivatives of the functions on the k'th row of A (bothersome!)
Homework Equations
Other than the expansion formula for a determinate I can't think of any.
The Attempt at a Solution
Seeing no other choice, I've tried proving this with induction. The problem is, I get lost in a maze of indexes and cannot seem to prove it.
for n=2 it's easy to show.
I assume it holds for n-1 and want to prove it for n.
That's as far as I got:
Here A[itex]^{ij}[/itex] is the minor of A at ij.
F'(x) = (detA)' = [itex](\sum_{k=1}^{n}f_{k1}detA^{k1})' = \sum_{k=1}^{n}[f'_{k1}detA^{k1} + f_{k1}(detA^{k1})'][/itex]
Now if we use the notation Fij(x) = det(A[itex]^{ij})[/itex], we get according to the induction assumtion:
= [itex]\sum_{k=1}^{n}[f'_{k1}detA^{k1} + f_{k1}\sum_{k=1}^{n}detA_{k}^{k1}][/itex]
And that's where I've pretty much given up. I doubt I'm on the right track.
Does anyone see a way out?
Thanks a lot! Tomer