How to prove that a system of linear equations doesn't exist for the solutions set $\{(a, a^2, b)|a,b \in \mathbb{R}\}$? I need to prove that a system of linear equations doesn't exist over $\mathbb{R}$ for the solutions set $\{(a, a^2, b)|a,b \in \mathbb{R}\}$. But isn't for example this system an example of when this does happen (for $x = y = 1$):
$$\begin{cases}
x+y=2 \\
2x+2y=4 \\
2x+y=3.
\end{cases}$$
 A: A non empty solution set  $X \subset \mathbb{R}^n$ to a set of linear equations is a sub-vectorspace of $\mathbb{R}^n$. In the case given let $X = \{(a,a^2,b) \mid a,b \in \mathbb{R} \}$. Consider the element $(1,1,0) \in X$ then if $X$ were a subspace of $\mathbb{R}^3$, certainly $2(1,1,0) \in X$. But if there are $a,b \in \mathbb{R}$ such that $2(1,1,0) = (a,a^2,b)$ then we must have $a = 2$ and $a = \sqrt{2}$ by considering the first two coordinates. So $X$ is not a subspace and so there is no set of linear equations for which $X$ is a solution.
A: Every non-trivial equation ($a_i \ne 0$) of a linear system of three variables describes an affine plane of $\mathbb{R}$:
$$
a_{i1} x_1 + a_{i2} x_2 + a_{i3} x_3 = b_i \iff \\
(a_{i1},a_{i2}, a_{i3}) \cdot (x_1,x_2,x_3) = b_i \iff \\
a_i \cdot x = b_i \iff \\
\frac{a_i}{\lVert a_i\rVert} \cdot x = \frac{b_i}{\lVert a_i\rVert} \iff \\
n_i \cdot x = d_i
$$
where $n_i$ is a unit normal vector of the $i$-th plane and $d_i$ is the (signed) distance of the plane from the origin.
A solution of a system must be element of the intersection of those planes.
So you can get, and thus model, either


*

*a plane, 

*a line, 

*a single point or 

*no solution (empty set)


These are the possible intersection sets of a non-empty set of planes in $\mathbb{R}^3$.
Your set 
\begin{align}
S 
&= \{ (a, a^2, b) \mid a, b \in \mathbb{R} \} \\
&= \{ (x, y, z) \in \mathbb{R}^3 \mid -x^2 + y = 0 \}
\end{align}
does not fit any of these. Here is a visualization for it:

A: This is another answer that was provided in our course. It doesn't involve knowledge of subspaces or any advanced notions at all. 
Let $S=\{(a,a^2,b)|a,b \in \mathbb{R}\}$. We have 3 unknowns, le them me $x,y,z$. 
Let $kx+ly+mz=r$ (*), $k,l,m,r \in \mathbb{R}$ be an equation in the system. Then for all $a,b, \in \mathbb{R}$: $ka+la^2+mb=r$.
If we choose, for example, $a=0$, $b=1$ then $m=r$.
If we choose $a=0$, $b=2$, then $2m=r$. IT follows then, that $m=r=0$, therefore the the equation (*) is of form $ka+la^2=0$. 
If we choose $a=1$, and then $a=-1$, then $k+l=0$ and $-k+l=0$, therefore $k=l=0$.
Therefore all equations in our system are of form $0x+0y+0z=0$ so the set of all solutions is essentially $\mathbb{R^3}$. 
But $S \neq \mathbb{R^3}$ because for example (1,2,1) is not of form $(a,a^2,b)$ because $1^2 \neq 2$.
