Let us compute the determinant of the jacobian transformation between the global x, y and a local L_{1}, L_{2}, L_{3} coordinate frame. One notices immediately that the problem is degenerate. That is why, we introduce a new coordinate z as a linear combination of L_{1}, L_{2}, L_{3} i. e. z = L_{1} + L_{2} + L_{3}. Note that z is not an independent coordinate and has a constant value equal 1. After taking into account relations Eq. (3) we find the jacobian matrix in the form
J(L_{1}, L_{2}, L_{3}) ≡  ∂(x, y, z)/∂(L_{1}, L_{2}, L_{3}) =  ( 
∂x/∂L_{1} ∂x/∂L_{2} ∂x/∂L_{3} ∂y/∂L_{1} ∂y/∂L_{2} ∂y/∂L_{3} ∂z/∂L_{1} ∂z/∂L_{2} ∂z/∂L_{3} 
)  =  ( 

) 
Furthermore, we have the relation between the jacobian and an element area
where Δ denotes the area of a triangle which is based on vertices (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}). And finally, we obtain the coordinates transformation rule
The relation between the gradient operator Δ in cartesian and in new coordinates is given by:
[∂/∂x, ∂/∂y] =  [∂L_{1}/∂x ∂/∂L_{1} + ∂L_{2}/∂x ∂/∂L_{2} + ∂L_{3}/∂x ∂/∂L_{3}, 
∂L_{1}/∂y ∂/∂L_{1} + ∂L_{2}/∂y ∂/∂L_{2} + ∂L_{3}/∂y ∂/∂L_{3}] 
[∂/∂x, ∂/∂y] =  1/(2Δ) [∂/∂L_{1}, ∂/∂L_{2}, ∂/∂L_{3}]  ( 

)  = [∂/∂L_{1}, ∂/∂L_{2}, ∂/∂L_{3}]  T 
where L_{k} = (a_{k}x + b_{k}y + c_{k})/(2Δ) (k = 1, 2, 3) and a_{1} = y_{2}  y_{3}, b_{1} = x_{3}  x_{2}, c_{1} = x_{2}y_{3}  x_{3}y_{2}, the rest of coeficients is obtained by cyclic permutation of indices 1, 2 and 3.