Rudin 9.19 Implicit Function Theorem Exercise 


*Show that the system of equations
$$
\begin{array}{r}
3 x+y-z+u^{2}=0 \\
x-y+2 z+u=0 \\
2 x+2 y-3 z+2 u=0
\end{array}
$$
can be solved for $x, y, u$ in terms of $z$; for $x, z, u$ in terms of $y$; for $y, z, u$ in terms of $x$; but not for $x, y, z$ in terms of $u$.


I am trying to show that this system can be solved for $x,y,u$ in terms of $z$. Does this need to have a solution set for every single $z\in\mathbb{R}$? I am thinking about applying the implicit function theorem however this would require checking that the derivative is non-invertible.
If I view this function as $f:\mathbb{R}^3\times \mathbb{R}\to \mathbb{R^3} $
$$f(x,y,z,u)=[3x+y-z+u^2, x-y+2z+u, 2x+2y-3z+2u]$$
I get that the total derivative w.r.t. (x,y,u) is
$$\left[\begin{array}{ccc}
3 & 1 & 2 u \\
1 & -1 & 1 \\
2 & 2 & 2
\end{array}\right]$$
Would this need to be invertible for every $u$ for the implicit function theorem to find a function in terms of $z$ that can solve the system of equations?
 A: Adding the last two equations and subtracting the first yields $3u-u^2=0$, then either $u=0$ or $u=3$. Therefore, unless u has one of these values there is no solution. so the system can not be solved for $x,y,z$ in terms of $u$. If one of these two equations holds we can solve just the last two equations for any two of the variables $x,y,z$ in terms of the third one. the remaining equations will be automatically satisfied. For instance
$$x= - \frac z 4 ,\,\,y = \frac{7z}{4},\,\, u=0;\,\, x= - \frac{9+z}{4},\,\, y = \frac{3+7z}{4},\,\, u=3.$$
Also, we could have
$$x= -\frac{y}{7},\,\, z = \frac{4y}{7},\,\,u=0;\,\, x = \frac{60+4y}{7},\,\, z= \frac{4y-3}{7},\,\, u=3.$$
Finally, we also could have
$$y = -7x,\,\, z=-4x,\,\, u=0;\,\, y =\frac{7x-60}{4},\,\, z= 9-4x,\,\, u=3.$$
Note that the matrix of the derivative of the transformation
$$f(x,y,z,u) = 
\begin{pmatrix}
3x+y-z+u^2 \\
 x-y+2z+u \\
2x+2y-3z+2u
 \end{pmatrix}$$
is
$$f'(x,y,z,u) = 
\begin{pmatrix}
3&1& -1&2u \\
 1&-1&2&1 \\
2&2&-3&2&
 \end{pmatrix}$$
and any $3 \times3$ submatrix containing the least column is invertible when $u=0$ or $u=3$. However, the first three columns of this matrix do not form an invertible matrix.
