1. **State the problem:** Find the Jacobian determinant $\frac{\partial(u,v,w)}{\partial(x,y,z)}$ where $$u = x^2 - 2y, \quad v = x + y + z, \quad w = x - 2y + 3.$$ 2. **Recall the formula for the Jacobian:** The Jacobian matrix $J$ is given by $$J = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{bmatrix}$$ The Jacobian determinant is $\det(J)$. 3. **Calculate each partial derivative:** - For $u = x^2 - 2y$: - $\frac{\partial u}{\partial x} = 2x$ - $\frac{\partial u}{\partial y} = -2$ - $\frac{\partial u}{\partial z} = 0$ - For $v = x + y + z$: - $\frac{\partial v}{\partial x} = 1$ - $\frac{\partial v}{\partial y} = 1$ - $\frac{\partial v}{\partial z} = 1$ - For $w = x - 2y + 3$: - $\frac{\partial w}{\partial x} = 1$ - $\frac{\partial w}{\partial y} = -2$ - $\frac{\partial w}{\partial z} = 0$ 4. **Form the Jacobian matrix:** $$J = \begin{bmatrix} 2x & -2 & 0 \\ 1 & 1 & 1 \\ 1 & -2 & 0 \end{bmatrix}$$ 5. **Calculate the determinant:** $$\det(J) = 2x \begin{vmatrix}1 & 1 \\ -2 & 0\end{vmatrix} - (-2) \begin{vmatrix}1 & 1 \\ 1 & 0\end{vmatrix} + 0 \begin{vmatrix}1 & 1 \\ 1 & -2\end{vmatrix}$$ Calculate each minor: - $\begin{vmatrix}1 & 1 \\ -2 & 0\end{vmatrix} = (1)(0) - (1)(-2) = 0 + 2 = 2$ - $\begin{vmatrix}1 & 1 \\ 1 & 0\end{vmatrix} = (1)(0) - (1)(1) = 0 - 1 = -1$ - The last term is multiplied by zero, so it is zero. 6. **Substitute back:** $$\det(J) = 2x \cdot 2 - (-2) \cdot (-1) + 0 = 4x - 2$$ 7. **Final answer:** $$\boxed{\det\left(\frac{\partial(u,v,w)}{\partial(x,y,z)}\right) = 4x - 2}$$ This is the Jacobian determinant of the transformation.