# Find basis of solutions of this linear system Supposed to find basis of the subspace of vector spac

Find basis of solutions of this linear system
Supposed to find basis of the subspace of vector space ${\mathbb{R}}^{3}$ of solutions of this linear system of equations:
$y=\left\{\begin{array}{l}{x}_{1}+2{x}_{2}-{x}_{3}=0\\ 2{x}_{1}+7{x}_{2}-2{x}_{3}=0\\ -{x}_{1}+3{x}_{2}+{x}_{3}=0\end{array}$
I solve this system and I got: ${x}_{1}={x}_{3}$ and ${x}_{2}=0$
$\stackrel{\to }{x}=\left[\begin{array}{c}{x}_{1}\\ 0\\ {x}_{1}\end{array}\right]={x}_{1}\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+0\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
Is the basis: $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$?
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thatuglygirlyu
By adding the first and the third equation we get $5{x}_{2}=0$ which implies ${x}_{2}=0$. If we now plug ${x}_{2}=0$ into remaining equations, we see, that they are all multiples of ${x}_{1}-{x}_{2}=0$.
Hence all solutions are of the form $\left({x}_{1},0,{x}_{1}\right)$. I.e., the solutions form the subspace $\left\{\left({x}_{1},0,{x}_{1}\right);{x}_{1}\in \mathbb{R}\right\}$.
Every vector in this subspace is a multiple of $\left(1,0,1\right)$, which means that vector $\left(1,0,1\right)$ generates the subspace.
Hence this subspace is one-dimensional and basis consists of a single vector $\left(1,0,1\right)$.
Of course, we could take any non-zero multiple of $\left(1,0,1\right)$ instead. For example, vector $\left(-2,0,-2\right)$ generates the same subspace.