Exercise 1.1.29

Fact: For any three vectors $u,v,w$ in the plane, some combination $cu+dv+ew$ is the zero vector (beyond the obvious $c=d=e=0$ ). So if there is one combination $Cu+Dv+Ew$ that produces $b$, there will be many more- just add $c,d,e$ or $2c,2d,2e$ to the particular solution $C,D,E$.

The example has $3u-2v+w=3(1,3)-2(2,7)+1(1,5)=(0,0).$ It also has $-2u+1v+0w=b=(0,1).$ Adding gives $u-v+w=(0,1).$ In this case $c,d,e$ equal $3,-2,1$ and $C,D,E=-2,1,0$.

Could another example have $u,v,w$ that could NOT combine to produce $b$ ? Yes. The vectors $(1,1),(2,2),(3,3)$ are on a line and no combination produces $b$. We can easily solve $cu+dv+ew=0$ but not $Cu+Dv+Ew=b$