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

${\displaystyle {\begin{matrix}2x&+5y&+2z&&-v&=&0\\\,3x&-4y&&+u&+2v&=&0\\\,,4x&&-2z&+2u&&=&0\,.\end{matrix}}}$
over ${\displaystyle {}\mathbb {R} }$. Due to
the solution set ${\displaystyle {}L}$ is a linear subspace of ${\displaystyle {}\mathbb {R} ^{5}}$. We have described it explicitly in example as
${\displaystyle {\left\{u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}},}$
which also shows that the solution set is a vector space. With this description, it is clear that ${\displaystyle {}L}$ is in bijection with ${\displaystyle {}\mathbb {R} ^{2}}$, and this bijection respects the addition and also the scalar multiplication (the solution set ${\displaystyle {}L'}$ of the inhomogeneous system is also in bijection with ${\displaystyle {}\mathbb {R} ^{2}}$, but there is no reasonable addition nor scalar multiplication on ${\displaystyle {}L'}$). However, this bijection depends heavily on the chosen "basic solutions“ ${\displaystyle {}{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}}$ and ${\displaystyle {}{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}}$, which depends on the order of elimination. There are several equally good basic solutions for ${\displaystyle {}L}$.