Prove that the matrix equation below using as matrix whose determinant is the Jacobian.

Formula used:

$$\begin{gathered} du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy \\ dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy \end{gathered}$$

Jacobian transformation:

$J=J\left(\frac{u,v}{x,y}\right)=\frac{\partial \left(u,v\right)}{\partial \left(x,y\right)}=\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right)$

Consider the following analytic function:

f(z) = u(x,y) + iv(x,y)

The above analytic function can also be written as follows:

w = f(z) = u(x,y) + iv(x,y)

The above analytic function f(z), defines a transformation from the variables x,y to the variables u, v then the Jacobian transformation is shown below:

$J=J\left(\frac{u,v}{x,y}\right)=\frac{\partial \left(u,v\right)}{\partial \left(x,y\right)}=\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right)$

Consider the matrix equation as shown below:

$J\left(\begin{array}{c}dx\\ dy\end{array}\right)=\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right)\left(\begin{array}{c}dx\\ dy\end{array}\right)$

$=\left(\begin{array}{cc}\frac{\partial u}{\partial x}dx+& \frac{\partial u}{\partial y}dy\\ \frac{\partial v}{\partial x}dx+& \frac{\partial v}{\partial y}dy\end{array}\right)$

...... (1)

Observe equation (1) to obtain:

$$\begin{gathered} du=\frac{du}{dx}dx+\frac{\partial u}{\partial y}dy \\ dv=\frac{dv}{dx}dx+\frac{\partial v}{\partial y}dy \end{gathered}$$

...... (2)

From (1) and (2) obtain:

$J\left(\begin{array}{c}dx\\ dy\end{array}\right)=\left(\begin{array}{c}du\\ dv\end{array}\right)$ ...... (3)

HereJ is a matrix whose determinant is the Jacobian.

The transpose of the Jacobian matrix is shown below:

$J\text{'}=\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial v}{\partial x}\\ \frac{\partial u}{\partial y}& \frac{\partial v}{\partial y}\end{array}\right)$

Multiply both sides of the matrix equation (3) by J^T .

$J.J^T.\left(\begin{array}{c}dx\\ dy\end{array}\right)=J^T\left(\begin{array}{c}du\\ dv\end{array}\right)$

Taking left-hand side of equation as follows:

$\begin{array}{rcl}J.J^T.\left(\frac{dx}{dy}\right)& =& \left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right).\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial v}{\partial x}\\ \frac{\partial u}{\partial y}& \frac{\partial v}{\partial y}\end{array}\right)\left(\begin{array}{c}dx\\ dy\end{array}\right)\\ & =& \left(\begin{array}{cc}{\left(\frac{\partial u}{\partial x}\right)}^2+{\left(\frac{\partial u}{\partial y}\right)}^2& 0\\ 0& {\left(\frac{\partial v}{\partial x}\right)}^2+{\left(\frac{\partial u}{\partial y}\right)}^2\end{array}\right)\left(\begin{array}{c}dx\\ dy\end{array}\right)\end{array}$

$\begin{array}{rcl}& & J^T.\left(\frac{du}{dv}\right)\end{array}$ : Right hand side part of equation (4).

Therefore, the matrix equation $\begin{array}{rcl}J\left(\frac{dx}{dy}\right)& =& \left(\frac{du}{dv}\right)\end{array}$ is verified.