Jacobian matrix of a variable transformation

[AxiomOfChoice]

Suppose I am changing variables from (x,y) to (s,t), where


\begin{align*}
s & = \frac 12 (x+y),\\
t & = y - x
\end{align*}


According to Wikipedia, if I want to see how the measure dx dy changes, I need to compute the Jacobian matrix J associated with this variable transformation and take its determinant. It will then follow that dx dy = \det J ds dt. The Jacobian matrix takes the form


\begin{bmatrix}
\partial x / \partial s & \partial x / \partial t \\
\partial y / \partial s & \partial y / \partial t
\end{bmatrix}
=
\begin{bmatrix}
1 & -1/2 \\
1 & 1/2
\end{bmatrix}


Is it just coincidence that the matrix J is identical to the matrix of the transformation; i.e., the matrix that shows up in the identity


\begin{bmatrix}
x \\ y
\end{bmatrix} =
\begin{bmatrix}
1 & -1/2 \\
1 & 1/2
\end{bmatrix}
\begin{bmatrix}
s \\ t
\end{bmatrix}

[chiro]

Suppose I am changing variables from (x,y) to (s,t), where


\begin{align*}
s & = \frac 12 (x+y),\\
t & = y - x
\end{align*}


According to Wikipedia, if I want to see how the measure dx dy changes, I need to compute the Jacobian matrix J associated with this variable transformation and take its determinant. It will then follow that dx dy = \det J ds dt. The Jacobian matrix takes the form


\begin{bmatrix}
\partial x / \partial s & \partial x / \partial t \\
\partial y / \partial s & \partial y / \partial t
\end{bmatrix}
=
\begin{bmatrix}
1 & -1/2 \\
1 & 1/2
\end{bmatrix}


Is it just coincidence that the matrix J is identical to the matrix of the transformation; i.e., the matrix that shows up in the identity


\begin{bmatrix}
x \\ y
\end{bmatrix} =
\begin{bmatrix}
1 & -1/2 \\
1 & 1/2
\end{bmatrix}
\begin{bmatrix}
s \\ t
\end{bmatrix}


Do you understand what the physical interpretation of the Jacobian is?

[HallsofIvy]

No, it is not coincidence. Since this is a linear transformation, its coefficients measure how much the measurement is stretched or contracted along the axes. And that determines how you will measure area.