Linear inhomogeneous system of equations/Elimination/2x+5y+2z-v, 3x-4y+u+2v, 4x -2z+2u/Example

We have a look at the homogeneous version of example, so we consider the homogeneous linear system

${\begin{matrix}2x&+5y&+2z&&-v&=&0\\\,3x&-4y&&+u&+2v&=&0\\\,,4x&&-2z&+2u&&=&0\,.\end{matrix}}$

over ${}\mathbb {R}$. Due to fact, the solution set ${}L$ is a linear subspace of ${}\mathbb {R} ^{5}$. We have described it explicitly in example as

${\left\{u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}},$

which also shows that the solution set is a vector space. With this description, it is clear that ${}L$ is in bijection with ${}\mathbb {R} ^{2}$, and this bijection respects the addition and also the scalar multiplication (the solution set ${}L'$ of the inhomogeneous system is also in bijection with ${}\mathbb {R} ^{2}$, but there is no reasonable addition nor scalar multiplication on ${}L'$). However, this bijection depends heavily on the chosen "basic solutions“ ${}{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}$ and ${}{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}$, which depends on the order of elimination. There are several equally good basic solutions for ${}L$.