Let $J(x,y)=x^2 -xy +y^2+3x-2y+1$ be a function defined on $E = R^2$.
Gradient : $ (2x -y +3, 2y-x-2)$
Hessian : $\binom{2 \ 0}{ 0 \ 2 }.$
the value of $a$ is : $y=3+2x \implies 2*(2y-x-2) +2x =4y-2x-4 \implies y=0.5x -1$
So i'm lost $a=\frac{5}{4};\frac{-14}{4}$??
I can not calculate the Hessian and deduce a local minimum point and is this a strict minimum, thanks
It sounds like you're looking for a stationary point of this function and are trying to figure out what kind of stationary point it is. However, you've run into a few problems.
1) To find the stationary point, you need to solve $\boldsymbol\nabla J = 0$. You've got the correct expression for the gradient, so now you just need to solve the system \begin{eqnarray} 3 + 2x - y &=& 0\\ -2-x+2y &=& 0 \end{eqnarray} whose solution is not $(x,y) = (5/4,-14/4)$.
2) To determine what kind of stationary point, you need to first evaluate the Hessian ($\boldsymbol\nabla\boldsymbol\nabla J$) at that point. You might want to check your calculation on that one, as it is not $[(2,0)(0,2)]$ (though you are correct that it is constant).
3) Once you have the Hessian, since this is a function of two variables, you simply need to check its determinant and the sum of its diagonal entries. A negative determinant is a saddle point; a positive determinant and positive sum of diagonal entries is a local minimum; and a positive determinant and negative sum of diagonal entries is a local maximum.