$\text{If } f(x,y,z)=\begin{bmatrix} x^2&y^2&z^2\\ x&y&z\\ 1&1&1 \end{bmatrix},\quad \text{prove that} fx + fy + fz = 0$
I have started the initial differential, however am not sure of the next step.
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1$\begingroup$ What space are you mapping into? $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ (so your question is using the determinant) or $f:\mathbb{R}^3 \rightarrow Mat_3(\mathbb{R}$) (the $3\times 3$ matrices with real entries)? $\endgroup$ – John Habert Feb 3 '14 at 20:38
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$\begingroup$ @ Jonn Talamantes: by $fx +fy + fz$ do you mean what is usually written $f_x + f_y + f_z$, where $f_x = \dfrac{\partial f}{\partial x}$ etc? $\endgroup$ – Robert Lewis Feb 3 '14 at 21:17
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$\begingroup$ @Robert Lewis: I just posted the same time an answer with the same comment. I think you are right, he must have meant that, since it is the only way the equation will be fulfilled. $\endgroup$ – Zoltan Zimboras Feb 3 '14 at 21:24
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Hint.
I think you wrote down the question with a bit misleading notation, as also John Habert pointed out. The only way it makes sense and your equation $fx+fy+fz=0$ is fulfilled, if you mean this
$$ f(x,y,z) = \text{det} \begin{bmatrix} x^2&y^2&z^2\\ x&y&z\\ 1&1&1 \end{bmatrix},$$
and instead of $fx$, $fy$, and $fz$ we have $f_x$, $f_y$, and $f_z$ - where these latter denote partial derivatives, $f_x=\tfrac{\partial f}{\partial x}$, $f_y=\tfrac{\partial f}{\partial y}$, and $f_z=\tfrac{\partial f}{\partial z}$.
Calculating the determinant, we get $f(x,y,z)=x^2y+y^2z +z^2x - z^2 y -y^2 x -x^2z$. Now obtain $f_x$, $f_y$ and $f_z$ by differentiating this expression wrt $x$, $y$ and $z$, respectively. This should be sufficient to get the answer.
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1$\begingroup$ This is how I interpreted the question as well but wanted clarification to be sure. $\endgroup$ – John Habert Feb 3 '14 at 21:49
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$\begingroup$ Yes, your comment also helped me interpreting it! $\endgroup$ – Zoltan Zimboras Feb 3 '14 at 21:57
