Proving that $\frac{D(s_1 , s_2)}{D(a,b)} \frac{D(a,b)}{D(x,y)} = \frac{D(s_1 ,s_2)}{D(x,y)}$ Using Lagrangian discribtion for incompressible fluid .
$$\frac{D(x,y)}{D(a,b)}=\begin{vmatrix}
\frac{\partial x}{\partial a} & \frac{\partial y}{\partial a}  \\ 
\frac{\partial x}{\partial b} & \frac{\partial y}{\partial b}  \\ 
\end{vmatrix}=1 $$
Prove that : $$\frac{D(s_1 , s_2)}{D(a,b)} \frac{D(a,b)}{D(x,y)} = \frac{D(s_1 ,s_2)}{D(x,y)}$$ 
Which implies that $$\begin{vmatrix}
\frac{\partial s_1}{\partial a} & \frac{\partial s_2}{\partial a}  \\ 
\frac{\partial s_1}{\partial b} & \frac{\partial s_2}{\partial b}  \\ 
\end{vmatrix} \begin{vmatrix}
\frac{\partial a}{\partial x} & \frac{\partial b}{\partial x}  \\ 
\frac{\partial a}{\partial y} & \frac{\partial b}{\partial y}  \\ 
\end{vmatrix}=\begin{vmatrix}
\frac{\partial s_1}{\partial x} & \frac{\partial s_2}{\partial x}  \\ 
\frac{\partial s_1}{\partial y} & \frac{\partial s_2}{\partial y}  \\ 
\end{vmatrix}$$
which implies that $$(\frac{\partial s_1}{\partial a} \frac{\partial s_2}{\partial b} -\frac{\partial s_2}{\partial a} \frac{\partial s_1}{\partial b})(\frac{\partial a}{\partial x} \frac{\partial b}{\partial y}-\frac{\partial b}{\partial x} \frac{\partial a}{\partial y})=2\frac{\partial s_1}{\partial x}\frac{\partial s_2}{\partial y}-\frac{\partial s_2}{\partial x}\frac{\partial s_1}{\partial y}$$
What is wrong ? 
It is not supposed to get 2 .
And the other question is if $$\frac{D(x,y)}{D(a,b)}=1$$
Then $$\frac{D(a,b)}{D(x,y)}=1$$
 A: I believe you have made a mistake, in that the following matrices are actually the correct matrices:
$\large\frac{D(x,y)}{D(a,b)}=\large\begin{vmatrix}
\frac{\partial x}{\partial a} & \frac{\partial x}{\partial b}  \\ 
\frac{\partial y}{\partial a} & \frac{\partial y}{\partial b}  \\ 
\end{vmatrix} \quad \frac{D(s_1,s_2)}{D(a,b)}=\large\begin{vmatrix}
\frac{\partial s_1}{\partial a} & \frac{\partial s_1}{\partial b}  \\ 
\frac{\partial s_2}{\partial a} & \frac{\partial s_2}{\partial b}  \\ 
\end{vmatrix} \quad \frac{D(a,b)}{D(x,y)}=\large\begin{vmatrix}
\frac{\partial a}{\partial x} & \frac{\partial a}{\partial y}  \\ 
\frac{\partial b}{\partial x} & \frac{\partial b}{\partial y}  \\ 
\end{vmatrix}$
Now, 
$$\frac{D(s_1,s_2)}{D(a,b)}\frac{D(a,b)}{D(x,y)} $$
$$=\large\begin{vmatrix}
\frac{\partial s_1}{\partial a} & \frac{\partial s_1}{\partial b}  \\ 
\frac{\partial s_2}{\partial a} & \frac{\partial s_2}{\partial b}  \\ 
\end{vmatrix}\large\begin{vmatrix}
\frac{\partial a}{\partial x} & \frac{\partial a}{\partial y}  \\ 
\frac{\partial b}{\partial x} & \frac{\partial b}{\partial y}  \\ 
\end{vmatrix} $$
$$=\det\left(\large\begin{bmatrix}
\frac{\partial s_1}{\partial a} & \frac{\partial s_1}{\partial b}  \\ 
\frac{\partial s_2}{\partial a} & \frac{\partial s_2}{\partial b}  \\ 
\end{bmatrix}\right)\det\left(\large\begin{bmatrix}
\frac{\partial a}{\partial x} & \frac{\partial a}{\partial y}  \\ 
\frac{\partial b}{\partial x} & \frac{\partial b}{\partial y}  \\ 
\end{bmatrix}\right)$$
$$= \det\left(\large\begin{bmatrix}
\frac{\partial s_1}{\partial a} & \frac{\partial s_1}{\partial b}  \\ 
\frac{\partial s_2}{\partial a} & \frac{\partial s_2}{\partial b}  \\ 
\end{bmatrix}\cdot\large\begin{bmatrix}
\frac{\partial a}{\partial x} & \frac{\partial a}{\partial y}  \\ 
\frac{\partial b}{\partial x} & \frac{\partial b}{\partial y}  \\ 
\end{bmatrix}\right) \\ \text{using the identity det(A)det(B) = det(AB)}$$
$$=\det\left(\large\begin{bmatrix}\frac{\partial s_1}{\partial a}\frac{\partial a}{\partial x} + \frac{\partial s_1}{\partial b}\frac{\partial b}{\partial x} & \frac{\partial s_1}{\partial a}\frac{\partial a}{\partial y} + \frac{\partial s_1}{\partial b}\frac{\partial b}{\partial y} \\ \frac{\partial s_2}{\partial a}\frac{\partial a}{\partial x} + \frac{\partial s_2}{\partial b}\frac{\partial b}{\partial x} & \frac{\partial s_2}{\partial a}\frac{\partial a}{\partial y} + \frac{\partial s_2}{\partial b}\frac{\partial b}{\partial y}\end{bmatrix}\right)$$
$$=\det\left(\large\begin{bmatrix}\frac{\partial s_1}{\partial x} & \frac{\partial s_1}{\partial y} \\ \frac{\partial s_2}{\partial x} & \frac{\partial s_2}{\partial y}\end{bmatrix}\right) \\ \text{by the chain rule for partial derivatives}$$
$$=\large\begin{vmatrix}\frac{\partial s_1}{\partial x} & \frac{\partial s_1}{\partial y} \\ \frac{\partial s_2}{\partial x} & \frac{\partial s_2}{\partial y}\end{vmatrix}$$
$$=\frac{D(s_1,s_2)}{D(x,y)}$$ as required
