Proving that $\frac{D(s_1 , s_2)}{D(a,b)} \frac{D(a,b)}{D(x,y)} = \frac{D(s_1 ,s_2)}{D(x,y)}$

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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 $$\frac{D(a,b)}{D(x,y)}=1$$

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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