Proving an exhausting relation in Algebra.

[Tomer]

Homework Statement

Hello, thanks for reading.

So, here's the problem. I'm given n² functions f_ij(x), and the matrix A = (f_ij)
I get the definition of a new function: F(x) = detA

I need to prove, that F'(x) = $\sum^{n}_{k=1}det(A_{k})$
Where A_k 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$^{ij}$ is the minor of A at ij.

F'(x) = (detA)' = $(\sum_{k=1}^{n}f_{k1}detA^{k1})' = \sum_{k=1}^{n}[f'_{k1}detA^{k1} + f_{k1}(detA^{k1})']$

Now if we use the notation F_ij(x) = det(A$^{ij})$, we get according to the induction assumtion:

= $\sum_{k=1}^{n}[f'_{k1}detA^{k1} + f_{k1}\sum_{k=1}^{n}detA_{k}^{k1}]$

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

 

[micromass]

What about proving it with the definition of the determinant:

$$\det(A)=\sum_{\sigma\in S_n}{sgn(\sigma)f_{1,\sigma(1)}f_{2,\sigma(2)}...f_{n,\sigma(n)}}$$

This should be a lot easier...

 

[Tomer]

I now realize I forgot the "sign" element in the expansion of the determinate above (how terribly bothersome!).

What is this "sigma" notation?