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Question
Math
Posted 2 months ago

Let ff be a transformation from R3\mathrm{R}^{3} to R3\mathrm{R}^{3}. Its Jacobian matrix is given below.
J(f)=[1013000cos(y)1]
J(f)=\left[\begin{array}{ccc}
-1 & 0 & -1 \\
3 & 0 & 0 \\
0 & \cos (y) & 1
\end{array}\right]


Find the Jacobian determinant of ff.
J(f)=
|J(f)|=

\square
How will ff expand or contract space around the point (0,π,0)(0, \pi, 0) ?
Choose 1 answer:
(A) Leave it the same
(B) Expand it finitely
(C) Contract it finitely
(D) Contract it infinitely
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 2
The determinant of J(f)J(f) is calculated as follows: J(f)=1(0100)0(3100)+(1)(3cos(y)00)|J(f)| = -1 \cdot (0 \cdot 1 - 0 \cdot 0) - 0 \cdot (3 \cdot 1 - 0 \cdot 0) + (-1) \cdot (3 \cdot \cos(y) - 0 \cdot 0)
step 3
Simplifying the expression, we get J(f)=100+(1)(3cos(y))|J(f)| = -1 \cdot 0 - 0 + (-1) \cdot (3 \cdot \cos(y))
step 4
Substituting y=πy = \pi, we have cos(π)=1\cos(\pi) = -1, so J(f)=1(31)=3|J(f)| = -1 \cdot (3 \cdot -1) = 3
step 5
Since the determinant is positive and non-zero, the transformation ff will expand space around the point (0,π,0)(0, \pi, 0) finitely
B
Key Concept
Determinant of a Jacobian matrix
Explanation
The determinant of the Jacobian matrix at a point gives us information about the behavior of the transformation at that point. A positive determinant indicates an expansion, while a negative determinant indicates a contraction. The magnitude of the determinant suggests the extent of the expansion or contraction.

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