Jacobians and change of variables
Problem 11.384 · easy
Find the Jacobian \( \displaystyle \dfrac{\partial(x, y)}{\partial(u, v)} \) of the transformation \( \displaystyle x = - 4 u + 2 v \), \( \displaystyle y = - u - 2 v \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial u} \left(- 4 u + 2 v\right) & \frac{\partial}{\partial v} \left(- 4 u + 2 v\right)\\\frac{\partial}{\partial u} \left(- u - 2 v\right) & \frac{\partial}{\partial v} \left(- u - 2 v\right)\end{matrix}\right] = \left[\begin{matrix}-4 & 2\\-1 & -2\end{matrix}\right] \]The matrix of partial derivatives.✓ Proved
- \[ 10 \]Its determinant: x_u·y_v − x_v·y_u.✓ Proved
- |J| is the factor by which the transformation stretches small areas: dA = |J| du dv.
Answer \( 10 \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the image of a tiny square measured by the shoelace formula |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The problem asks for the Jacobian determinant, but the solution concludes with the absolute value |J|. While the final numerical answer is correct because the determinant is positive, the text implies that the Jacobian itself is defined as the absolute value, which is false; the Jacobian is the signed determinant, and the absolute value is only used for area scaling in integrals.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: fail (style) 2026-10-13 — [domain objection, downgraded to style] The problem asks for the Jacobian determinant, but the solution concludes with the absolute value |J|. While the final numerical answer is correct because the determinant is positive, the text implies that the Jacobian itself is defined as the absolute value, which is false; the Jacobian is the signed determinant, and the absolute value is only used for area scaling in integrals.qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly computes the Jacobian determinant and provides a standard, correct interpretation of its geometric meaning in the context of change of variables.gpt-oss:20b: pass 2026-10-13
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/jacobian, checked 2026-10-13 with SymPy 1.14.0.