Why is this system insoluble? I am trying to show the following simultaneous equations
$$3x+y-z+u^2 = 0$$
$$x-y+2z+u = 0$$
$$2x+2y-3z+2u = 0$$
can be solved for
(i) $x,y,u$ in $z$ (i.e. in terms of $z$), (ii) $x,z,u$ in $y$, (iii) $y,z,u$ in $x$, but NOT for (iv) $x,y,z$ in $u$. Also, for part (iv), if we express the system as a matrix equation in $x,y$ and $z$, the determinant is $0$, but how can one go from there to show the desired insolubility in terms of $u$? There could, for instance, be infinitely many solutions in terms of $u$.
 A: Adding the first equation to the second equation gives $4x + z + u^2 +u = 0$, and adding twice the second equation to the third equation gives $4x + z + 4u = 0$; together these equations imply $u^2 + u = 4u$, or $u = 3$ or $u = 0$. Thus the system of equations cannot be solved in terms of $u$ unless $u$ is either one of these two values. For the remaining cases, take $u = 0$. Using Gaussian elimination, an equivalent system of equations is
\begin{align*}
4x + z &= 0 \\
4y - 7z &= 0.
\end{align*}
It is clear that $x$ and $y$ can be written in terms of $z$, $x$ and $z$ in terms of $y$, and $y$ and $z$ in terms of $x$. This completes the proof.
A: If you are not solving for $u$, then your system is in fact linear; it is
$$\begin{bmatrix} 3 & 1 & -1 \\ 1 & -1 & 2 \\ 2 & 2 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -u^2 \\ -u \\ -2u \end{bmatrix}.$$
Systems given by the matrix on the left are not always solvable because the determinant of the matrix is zero. (I just calculated this on a computer, not by hand, but it's not hard to do by hand either.) This does not quite finish the problem, because the right hand sides are not arbitrary. However, it is possible to cook up three linearly independent right hand sides of that form, e.g. by taking $u=1,-1,2$. One of these must make the system unsolvable, for otherwise the matrix would be invertible.
