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Chapter 13: Problem 38

Evaluate the Jacobians \(J(u, v, w)\) for the following transformations. $$x=u+v-w, y=u-v+w, z=-u+v+w$$

Short Answer

Expert verified
$$ Answer: The value of the Jacobian determinant, \(J(u, v, w)\), is 0.

Step by step solution

01

Calculate partial derivatives

Find the partial derivatives of x, y, and z with respect to u, v, and w. $$\frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = 1, \quad \frac{\partial x}{\partial w} = -1$$ $$\frac{\partial y}{\partial u} = 1, \quad \frac{\partial y}{\partial v} = -1, \quad \frac{\partial y}{\partial w} = 1$$ $$\frac{\partial z}{\partial u} = -1, \quad \frac{\partial z}{\partial v} = 1, \quad \frac{\partial z}{\partial w} = 1$$
02

Form the Jacobian matrix

Arrange the partial derivatives in a 3x3 matrix to form the Jacobian matrix J. $$J(u, v, w) = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{pmatrix} = \begin{pmatrix} 1 & 1 & -1\\ 1 & -1 & 1\\ -1 & 1 & 1 \end{pmatrix}$$
03

Evaluate the determinant

Determine the determinant of the Jacobian matrix J to find the value of \(J(u, v, w)\). $$J(u, v, w) = \text{det}(J) = \text{det}\begin{pmatrix}1 & 1 & -1\\ 1 & -1 & 1\\ -1 & 1 & 1\end{pmatrix}$$ Expanding the determinant along the first row: $$\begin{aligned} J(u, v, w) &= 1 \cdot \text{det}\begin{pmatrix}-1 & 1\\ 1& 1\end{pmatrix} - 1 \cdot \text{det}\begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix} + (-1) \cdot \text{det}\begin{pmatrix}1 & -1\\ -1 & 1\end{pmatrix}\\ &=(-1) \cdot (1\cdot 1 - 1\cdot (-1)) + (-1)(1\cdot(1\cdot 1 - 1\cdot (-1) )\\ &=(-1) \cdot (2) - (-1)(2)\\ &=-2 + 2 = 0 \end{aligned}$$ Thus, \(J(u, v, w) = 0\).

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