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Suppose $f = \left( f_1, f_2, \ldots, f_{n-1} \right) : \mathbb{R}^{n} \mapsto \mathbb{R}^{n-1}$ is a $C^{2}$ function, then show that the symbolic determinant

\begin{align} \begin{vmatrix} \frac{\partial}{\partial x_{1}} &\frac{\partial f_{1}}{\partial x_{1}} &\frac{\partial f_{2}}{\partial x_{1}} &\cdots &\frac{\partial f_{n-1}}{\partial x_{1}} \\\\ \frac{\partial}{\partial x_{2}} &\frac{\partial f_{1}}{\partial x_{2}} &\frac{\partial f_{2}}{\partial x_{2}} &\cdots &\frac{\partial f_{n-1}}{\partial x_{2}} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ \frac{\partial}{\partial x_{n}} &\frac{\partial f_{1}}{\partial x_{n}} &\frac{\partial f_{2}}{\partial x_{n}} &\cdots &\frac{\partial f_{n-1}}{\partial x_{n}} \end{vmatrix} \end{align}

vanishes identically.

I have been trying to rack my brains thinking of various methods which can be used to solve the following problem, but I am getting nowhere, I am not particularly good at theoretical multivariable calculus, and hence might be missing some basic concept here. I would be thankful if someone could point out a direction for me to work through.

P.S. This problem is from the entrance examination, 2010 to the Graduate School at Chennai Mathematical Institute.

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Let $D$ denote the determinant in question. Observe: $$D = \sum_I \epsilon_{i_1i_2\dots i_n}\partial_{i_1}\partial_{i_2}f_1 \cdots \partial_{i_n}f_{n-1}$$ But, we sum over antisymmetric $i_1i_2\ldots i_n$ in the antisymmetric symbol against the symmetric $i_1i_2\ldots i_n$ appearing in the derivatives (mixed partials commute given the supposed differentiability). Thus, it is zero.

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That's all there to it?! I came up with a similar but infinitely cruder solution, however, I was not sure of this. Also, I was thinking there might be some rank-related solution also. Anyway, thanks. :-) –  Jayesh Badwaik Oct 26 '12 at 14:11
    
@James. Just for my clarification.This can be realised as a cross product right? –  Vishesh Oct 26 '12 at 14:23
1  
there is an $n$-ary cross-product on $\mathbb{R}^n$ given by a determinant. I suppose you could view the formula in that light. One way or the other you'll need either to use an argument like the one I gave or a property of determinants. But, be careful because one column is an operator whereas the others are functions. It's a hybrid expression so properties may not apply as you expect. Analogy: $A \cdot B = B \cdot A$ however $\nabla \cdot F \neq F \cdot \nabla$ (lhs is function, rhs is operator) –  James S. Cook Oct 26 '12 at 15:49

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