Suppose I want to prove the following identity:

$\nabla. (F \times G)= G.(\nabla \times F) - F.(\nabla \times G)$ for a vector on $\ \mathbb{R}^3$

I know that the "most correct" way to prove this is by invoking the del operator on the vector $\ F \times G$. However, I tried to use the pseudo determinant representation of $\ \nabla \times G$ and $\ \nabla . G$ which I thought would help me to prove these identities faster. However, I realised that some of the time this works, while for other times, this does not work.

Particularly, if you see $\nabla. (F \times G)$ as a determinant of a 3 x 3 matrix as in my working on the LHS, this actually works. However, if we see $\ G.(\nabla \times F)$ as such and apply properties of determinants, it does not seem to work. At least I feel that the step on applying an elementary row operation is really suspicious but I can't point how it is so...

Can anyone enlighten me on why this works sometimes, while it doesnt in other cases? Thanks!

Link to image for my working


1 Answer 1


Your row reduction is problematic because the differential operator is not commutative in the evaluation of the determinant.   They are not scalars after all, so row reductions of matrices containing operator elements can change the entire meaning of the matrix.

For example, usually $P_F\tfrac{\partial ~}{\partial y}R_G \neq R_G\tfrac{\partial ~}{\partial y}P_F $.

So the $R_1\leftrightarrow R_3$ exchange is not going to simply change the sign of the determinant.

What you've done is analogous to claiming$$\tfrac {\mathrm d ~~}{\mathrm d ~x}\big(f(x)g(x)\big)~{= f(x)\big(\tfrac{\mathrm d ~~}{\mathrm d ~x}g(x)\big)+\big(\tfrac{\mathrm d ~~}{\mathrm d ~x}f(x)\big)g(x) \\ = \tfrac{\mathrm d ~~}{\mathrm d ~x}\big(f(x)g(x)\big)+\tfrac{\mathrm d ~}{\mathrm d ~x}\big(f(x)g(x)\big)\\=2\tfrac{\mathrm d ~}{\mathrm d ~x}\big(f(x)g(x)\big)}$$

  • $\begingroup$ Ok thanks alot! $\endgroup$
    – HK Tan
    Commented Apr 20, 2018 at 7:50

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