Finding a Jacobian determinant of transformation 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).
 A: $\def\d{\partial}$
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.
$$\begin{align}
\left(\begin{array}\\\frac{\d x}{\d u} & \frac{\d x}{\d v} & \frac{\d x}{\d w}\\\frac{\d y}{\d u} & \frac{\d y}{\d v} & \frac{\d y}{\d w}\\\frac{\d z}{\d u} & \frac{\d z}{\d v} & \frac{\d z}{\d w} \end{array}\right) = \left(\begin{array}\\\frac23 & \frac{-2}3 & \frac13\\\frac23 & \frac{-2}3 & \frac{-1}3\\\frac23 & \frac23 & \frac13 \end{array}\right) 
\end{align}$$
