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$$x_1 − 2x_2 + x_3 = 0$$ $$2x_1 − 3x_2 + x_3 = 0$$

Find the basis of this. I got that $x_1 = x_2 = x_3$ but I am not sure what should I do with that. I also lowered it to lower echealon form and got $$[1, 0, -1]$$ $$ [0, 1, -1]$$ Is the basis like (1,0,-1),(0,1,-1), (0,0,-1)?

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Since you have a free variable $x_3=t \in \mathbb{R} $ then the solution vector is given by

$$(x_1,x_2,x_3)=(t,t,t)=t(1,1,1)$$

which means that the set $\left\{(1,1,1)\right\}$ is a basis for the subspace.

Note: I used the solution

$$x_1=x_2=x_3. $$

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