This resolution, correct?
"Find a basis of size and $U=\{{(x,y,z)}\in R^{3};3x-y-z=0\}$"$$$$My
response:$$3x-y-z=0\Longrightarrow y=3x-z$$$$(x,y,z)=(x,3y-z,z)$$Making
$x$ and $z$ equals zero, one at a time, we
have$$x=0\Longrightarrow(0,-z,z)$$$$z=0\Longrightarrow(x,3x,0)$$Adding
these results:$$(0,-z,z)+(x,3x,0)=z(0,-1,1)+x(1,3,0)$$Thus, a base
is$$\{(0,-1,1),(1,3,0)\}$$And the size is $$\text{dim}\;2$$ $$$$ This
resolution, correct?
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