lilmoore11p8

2022-07-07

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: $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$?

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.

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