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I want to find the Jacobian determinant of the transformation:

$$\begin{align} \left(\begin{array}\\x\\y\\z\end{array}\right) = \left(\begin{array}\\\frac23 & \frac{-2}3 & \frac13\\\frac23 & \frac{-2}3 & \frac{-1}3\\\frac23 & \frac23 & \frac13 \end{array}\right) \left(\begin{array}\\u\\v\\w\end{array}\right) \end{align}$$

What exactly does it want me to do? What does it mean by find the Jacobian determinant?

Note: Some of the values have been changed to not be cheating potentially, I hope that doesn't make the question impossible. I did change the values evenly though(take one off the top and give it to another for example).

Git Gud
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  • Do you know what the jacobian of a function is? – Git Gud May 17 '14 at 12:52
  • @GitGud I have used it for polar and spherical coordinates. From what I understand, it takes all the partial derivatives of specific things, but here I have no function?

    Do I just evaluate $M_{3\times3} * M_{3\times1}$ and find what $x$ is in terms of $u$?

     –  May 17 '14 at 12:57
  • See this wikipedia page. You need to identify your $F$. If you know what it is, the answer should follow easily. Do you know what it is? – Git Gud May 17 '14 at 13:23
  • @GitGud I perhaps do, $x = \frac13(2u-2v+w), y = \frac13(2u-2v-w), z = \frac13(2u + 2v + w)$? –  May 17 '14 at 13:33
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    If you understand exactly what you're saying, than that's it. I, myself, would rather write $F(u,v,w)=\left(\frac13(2u-2v+w), \frac13(2u-2v-w), \frac13(2u + 2v + w)\right)$. In the notation of the link you have $F_1=x, F_2=y, F_3=z$ and $n=3$. If you got this, please post an answer yourself so the question doesn't come up as answered. – Git Gud May 17 '14 at 13:37
  • @GitGud Could you possibly have a look at my answer and check that it is coherent? –  May 17 '14 at 14:30
  • It's fine.${{{}}}$ – Git Gud May 17 '14 at 15:02

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$\def\d{\partial}$ To find the determinant of the transformation matrix, we need to complete the Jacobian matrix:

$$\begin{align} \left(\begin{array}\\\frac{\d x}{\d u} & \frac{\d x}{\d v} & \frac{\d x}{\d w}\\\frac{\d y}{\d u} & \frac{\d y}{\d v} & \frac{\d y}{\d w}\\\frac{\d z}{\d u} & \frac{\d z}{\d v} & \frac{\d z}{\d w} \end{array}\right) \end{align}$$

For:

$$ {\bf A}=\begin{align} \left(\begin{array}\\x\\y\\z\end{array}\right) = \left(\begin{array}\\\frac23 & \frac{-2}3 & \frac13\\\frac23 & \frac{-2}3 & \frac{-1}3\\\frac23 & \frac23 & \frac13 \end{array}\right) \left(\begin{array}\\u\\v\\w\end{array}\right) \end{align}$$

$$\begin{align} \left(\begin{array}\\x\\y\\z \end{array}\right)=\left(\begin{array}\\\frac{2u}3- \frac{2v}3+ \frac{w}3\\\frac{2u}3 - \frac{2v}3 - \frac{w}3\\\frac{2u}3 +\frac{2v}3 +\frac{w}3 \end{array}\right) \end{align}$$

Since the above is linear, and already constructed in $(u,v,w)$ order, the original matrix is indeed the Jacobian Determinant. Note that $(u,v,w)$ is transformed by this matrix.

$$\begin{align} \left(\begin{array}\\\frac{\d x}{\d u} & \frac{\d x}{\d v} & \frac{\d x}{\d w}\\\frac{\d y}{\d u} & \frac{\d y}{\d v} & \frac{\d y}{\d w}\\\frac{\d z}{\d u} & \frac{\d z}{\d v} & \frac{\d z}{\d w} \end{array}\right) = \left(\begin{array}\\\frac23 & \frac{-2}3 & \frac13\\\frac23 & \frac{-2}3 & \frac{-1}3\\\frac23 & \frac23 & \frac13 \end{array}\right) \end{align}$$