Show transformation is locally invertible $$\begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix}= G\begin{pmatrix} u_1\\ u_2\\ u_3\\ \end{pmatrix}$$ with
\begin{align*}
x_1&=u_1+2 \cdot u_3 \\
x_2&=u_3 –2\cdot  u_1 \\
x_3&=u_1+u_2+u_3
\end{align*}
a)   Computer the derivate of this transformation 
For this I have made the matrix :
\begin{bmatrix} 1 & 0 & 2u3 \\ -2u1 & 0 & 1 \\ 1 & 1 & 1\end{bmatrix}
b)   Use the inverse function theorem to show that the transformation is locally invertible if $u_1u_2>0$
The derivative I get is $-1-4\cdot u_1 u_2$   but to satisfy the inverse function theorem all I need to show is that         $-1-4\cdot u_1 u_2$  not equal $0$.  Not sure how I can show that $u1u2 > 0$
 A: The derivative of that transformation should be a $3\times 3$ matrix, and when $u_1 u_2>0$ than the determinant will be not zero, so it is local invertible.
A: As written at the moment the map $G:\ \Bbb R^3\to\Bbb R^3$ is  a linear map with matrix
$$\left[\matrix{1&0&2\cr -2&0&1\cr 1&1&1\cr}\right]\ .$$
The determinant of this matrix is $-5\ne0$. It follows that $G$ is bijective and has an inverse in the sense of linear algebra. This inverse
$$G^{-1}:\quad \Bbb R^3\to\Bbb R^3, \qquad y\mapsto x=G^{-1}y$$
can be considered as a differentiable map in its own right.
$G$ being linear its derivative is  constant. To be precise: For any $x\in \Bbb R^3$ we have $dG(x)=G$. This is an immediate consequence of $G(x+h)=G(x)+G(h)$, where $x$ is momentarily fixed and $h$ is a variable tangent vector attached to $x$.
As $\det\bigl(d G(x)\bigr)=\det(G)=-5$ for all $x\in{\mathbb R}^3$ it follows that $G$ is at each $x$ locally invertible in the sense of multivariate calculus. Of course we already knew this, as $G$ is globally linear.
