Define a new set of coordinates $u, v, w$ in terms of $x, y, z$: $$u = x + xyz,\\ v = y + x\\ w = 2x + z + 3z^2 $$
Can the system be solved for $x, y,$ and $z$ in terms of $u, v, $and $w$ near $\begin{pmatrix} 0 \\ 0 \\ 0\\ \end{pmatrix}$. Justify your answer.
To my understanding, this question should be solved using the implicit function theorem. However, the past questions I have done has involved concrete numbers for $x, y, z$. In this case, how would I continue with $u, v, w$ as arbitrary coordinates and how should I substitute the column vector, as $x, y, z$ or as $u, v, w$
Define the $3\space C^1$ functions: $M(x,y,z,u,v,w) = x+xyz-u, N(x,y,z,u,v,w) = x+y-v, P(x,y,z,u,v,w) = 2x+3z^2+z - w$. Next, you calculate the determinant $\text{det}\begin{bmatrix} \dfrac{\partial M}{\partial x} & \dfrac{\partial M}{\partial y} & \dfrac{\partial M}{\partial z} \\ \dfrac{\partial N}{\partial x} & \dfrac{\partial N}{\partial y} & \dfrac{\partial N}{\partial z} \\ \dfrac{\partial P}{\partial x} & \dfrac{\partial P}{\partial y} & \dfrac{\partial P}{\partial z} \end{bmatrix}|_{(0,0,0,0,0,0)}=\text{det}\begin{bmatrix} 1+yz & xz & xy \\ 1 & 1 & 0 \\ 2 & 0 & 6z+1\end{bmatrix}|_{(0,0,0,0,0,0)}= \text{det}\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix} = 1 \neq 0$. Thus the implicit function theorem guarantees that you can solve for $x,y,z$ in terms of $u,v,w$ near the point $(0,0,0,0,0,0)$.