In C.H. Edward's Advanced Calculus of Several Variables in the Chapter III in Section 3 on Inverse and Implicit Mapping Theorems question #5 is given as follows:
3.5 Show that the equations
$$ \sin(x+z)+\log yz^2 = 0, \qquad e^{x+z}+yz=0 $$
implicitly define $z$ near $-1$ as a function of $(x,y)$ near $(1,1)$.
My trouble here is in understanding what is meant by the statement. If I am to consider the collection of points in $\mathbb{R}^3$ for which both $G_1(x,y,z)= \sin(x+z)+\log yz^2 = 0$ and $G_2(x,y,z)=e^{x+z}+yz=0$ then, if the Jacobian is rank two near $(1,1,-1)$ then we have two (independent) equations and three unknowns hence the implicit solution would have just one free parameter. That is, the implicit solution would seem to be a curve not a surface. Very well, for completeness, let us calculate the Jacobian:
$$ G'(x,y,z) = [\partial_xG|\partial_yG|\partial_zG] =
\left[ \begin{array}{ccc} \cos(x+z) & 1/y & \cos(x+z)+2/z \\
e^{x+z} & z & e^{x+z}+y \end{array}\right] $$
and evaluate at $(1,1,-1)$ for which clearly $G(1,1,-1)=(0,0)$
$$ G'(1,1,-1) =
\left[ \begin{array}{ccc} 1 & 1 & -1 \\
1 & -1 & 2 \end{array}\right] $$
It is clear to me that the matrix above has rank two. Moreover, we can apply the implicit mapping theorem to solve for $x,z$ as a functions of $y$, that is we could find $h_1,h_2$ functions for which $G(h_1(y),y,h_2(y))=(0,0)$ for $y$ near $1$. Or, we could solve for $x,y$ are functions of $z$, that is we could find $h_3,h_4$ such that $G(h_3(z),h_4(z),z)=(0,0)$ for $z$ near $-1$. Finally (thanks to Ted Shifrin's correction) we can also find solution exists for $z,y$ in terms of $x$ exists near $(1,1,-1)$.
In short, if I consider both equations at once, I don't see a surface parametrized by $x,y$ near $(1,1,-1)$.
Question: does Edwards intend us to consider the equations $G_1=0$ and $G_2=0$ separately? In that case, I can solve for two (distinct) surfaces implicitly given by $G_1(x,y,z)=0$ and $G_2(x,y,z)=0$ respective. The intersection of those surfaces would correspond to the curve which the implicit mapping theorem applied to $G = (G_1,G_2)$ revealed. Am I missing something here?
Best Answer
There's clearly a typo in the question. My copy of the book is in my office, so I can't check it now. However, you did miscalculate the formulas for the partial derivatives. I get the Jacobian matrix $$\begin{bmatrix} 1 & 1 & -1 \\ 1 & -1 & 2 \end{bmatrix}\,.$$ So since all three $2\times 2$ minors are nonzero, we can locally express this curve as a graph in any of the three ways.