Let \( f(x, y) = (x^2 - y^2, 2xy) \), where \( x>0, y>0 \). Let \( g \) be the inverse of \( f \) in a neighborhood of \( f(2, 1) \). Then the determinant of the Jacobian matrix of \( g \) at \( f(2, 1) \) is equal to (round off to TWO decimal places):
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Solution and Explanation
2y & 2x \end{bmatrix}. \] At \( (x, y) = (2, 1) \), \[ J_f = \begin{bmatrix} 4 & -2
2 & 4 \end{bmatrix}. \] Step 2: Determinant of \( J_f \). \[ \det(J_f) = (4)(4) - (-2)(2) = 16 + 4 = 20. \] Step 3: Determinant of \( J_g \). \[ \det(J_g) = \frac{1}{\det(J_f)} = \frac{1}{20} = 0.05. \] Step 4: Conclusion. The determinant of \( J_g \) is \( {0.05} \).
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