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$\phi=(\phi_2,\dots,\phi_n):\mathbb{R}^n\rightarrow\mathbb{R}^{n-1}$ is a $C^2$ function, we need to show that the synbolic determinant $$\begin{pmatrix}\partial/\partial x_1&\partial\phi_2/\partial x_1&\partial\phi_3/\partial x_1\\\partial/\partial x_3&\partial\phi_2 /\partial x_3&\partial\phi_3/\partial x_3\\\dots\\\partial/\partial x_n\dots&\dots&\partial \phi_n/\partial x_n\end{pmatrix}$$ vanishes identically.

First I would like to know what is symbolic determinant,I am not sure whether I have written the matrix properly,and I dont know how write determinant so I wrote matrix. Thank you for help.

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Hmmm... How does one compute the determinant of a non-square matrix? –  Haskell Curry Jan 21 '13 at 15:51
Hmm...I told that I dont know what is symbolc determinant, sorry I missed one row more. –  Une Femme Douce Jan 21 '13 at 15:53

1 Answer 1

up vote 1 down vote accepted

Symbolic means without numbers;

Your matrix is incorrect; you can't determine its determinant. However you might mean this one.

$$\det \begin{pmatrix}\partial\phi_2/\partial x_1&\partial\phi_3/\partial x_1\\\partial\phi_2 /\partial x_3&\partial\phi_3/\partial x_3\end{pmatrix}$$

Which "vanishes":

$$\partial\phi_2/\partial x_1 \cdot \partial\phi_3 /\partial x_3 - \partial\phi_3/\partial x_1 \cdot \partial\phi_2/\partial x_3 =\frac{\partial \phi_2 \cdot \partial \phi_3-\partial \phi_3 \cdot \partial \phi_2}{\partial x_1 \cdot \partial x_3}$$

*I see you edited your post; nevertheless the idea remains the same.

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