Variation of the determinant of a Jacobian I am following a derivation in a Calculus of Variation problem.
After introducing a one-parameter family of one-to-one mappings from $R^{2}$ to itself, $$z({x},\epsilon)$$, $x = (x_1,x_2)$, such that $$z({x},0) = x$$.
The mapping is used for a change of variable within an integral, hence the issue of computing the variation of the jacobian's determinant arises.
The author says that the obvious identity $$\delta \, det z_{a,b} = \delta z_{a,a}$$
Well as it often happens, it is not entirely obvious to some...Any help would be so appreciated, thanks

Hints: 


*

*Consider a differentiable function $f: \mathbb{R}^n \times \mathbb{R}\to \mathbb{R}^n$ [let us write $z=f(x,\epsilon)$] with the property that $f(\cdot , 0)={\rm id}_{\mathbb{R}^n}$ is the identity map.

*Define Jacobi matrix $A:=\frac{\partial f}{\partial x}$. Then $\left. A \right|_{\epsilon=0}={\bf 1}_{n\times n}$ is the identity matrix.

*One can argue in several ways that the sought-for identity must hold, e.g.
$$ \left.\frac{\partial}{\partial\epsilon} \det(A)  \right|_{\epsilon=0}
~=~\left. \det(A)^{-1} \frac{\partial}{\partial\epsilon} \det(A)  \right|_{\epsilon=0}
~=~\left.\frac{\partial}{\partial\epsilon}\ln\det(A)  \right|_{\epsilon=0}$$
$$~=~\left.\frac{\partial}{\partial\epsilon}{\rm tr}\ln(A) \right|_{\epsilon=0}
~=~\left.{\rm tr}(A^{-1}\frac{\partial}{\partial\epsilon}A) \right|_{\epsilon=0}
~=~\left.\frac{\partial}{\partial\epsilon}{\rm tr}(A) \right|_{\epsilon=0}. $$
