Let $f(x,y)=(x^2-y^2,2xy),$ where $x>0,y>0.$ Let $g$ be the inverse of $f$ in a neighbourhood of $f(2,1)$. Then the determinant of the Jacobian matrix of $g$ at $f(2,1)$ is equal to...............?
Solution: Let $f(x,y)=(f_1(x,y),f_2(x,y))$,where $f_1(x,y)=x^2-y^2$ and $f_2(x,y)=2xy$,then the Jocobian of the matrix is,$Jf=\begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix}=\begin{bmatrix} 2x & -2y \\ 2y & 2x \end{bmatrix}$,then $Jf(2,1)=\begin{bmatrix} 4 & -2 \\ 2 & 4 \end{bmatrix}$,$det(Jf(2,1))=det \begin{bmatrix} 4 & -2 \\ 2 & 4 \end{bmatrix}=20\neq 0$,then by Inverse function theorem $det (Jg(2,1))=det (Jf^{-1}(2,1))=\frac{1}{det((Jf(2,1)))}=1/20$
Is my Solution correct?Please suggest anyother method by which it can be tackled also check whether I applied the Inverse function theorem correctly
Mostly.
$f(2,1)=(3,4)$ so you want $\det(\mathrm J_g(3,4)) = \det(\mathrm J_{f^{-1}}(3,4))$
But otherwise, yes, that is: $\det(\mathrm J_f(2,1))^{-1}=1/20$.