System of linear equations - geometrical representation of solution

Question

So, if we're given the system: $$x+ay=1$$ $$-ay+z=a-1$$ $$2z=a$$ where $a\neq 0$, we can write it's solutions as: $$x=\frac{a}{2}$$ $$y=-\frac{1}{2}+\frac{1}{a}$$ $$z=\frac{a}{2}$$ or: $$(x,y,z)=\bigg\{\bigg(\frac{a}{2},-\frac{1}{2}+\frac{1}{a},\frac{a}{2}\bigg):\frac{a\in\mathbb{R}} {0}\bigg\}$$ I also wrote it as: $$(x,y,z)=\bigg\{\bigg(0,-\frac{1}{2},0\bigg)+a\cdot\bigg(\frac{1}{2},0,\frac{1}{2}\bigg)+\frac{1}{a}\cdot\bigg(0,1,0\bigg)\bigg \}$$ How can I understand this solution geometrically? If, instead of $\frac{1}{a}$, we had some other real number $b$, this would be a linear combination of two vectors in $\mathbb{R}^3$ shifted by vector $(0,-\frac{1}{2},0)$, which is a plane. But, in this case, we have two vectors in a linear combination, which is defined only by one real number $a$.

Is this solution even a vector space? For which parameter $a$ can we get a $0$-vector?
Thank you for your time.

Answer 1

It is an Hyperbola in the plane $x=z$ with center at $(0,-1/2,0)$.

So it is not a vector space and we cannot have the $0$ vector for any value of $a$.