Let us write the vector (1,−2,5) as a linear combination of the vectors (1,1,1),(1,2,3)
and (2,−1,1).
Let
(1,−2,5)=a(1,1,1)+b(1,2,3)+c(2,−1,1)=(a+b+2c,a+2b−c,a+3b+c).
It follows that we get the system
⎩⎨⎧a+b+2c=1a+2b−c=−2a+3b+c=5
which is equivalent after subtracting from the second row the first row, and from the third row the second row to the system
⎩⎨⎧a+b+2c=1b−3c=−3b+2c=7
and hence the last system is equivalent after subtracting from the third row the second row to the system
⎩⎨⎧a+b+2c=1b−3c=−35c=10
It follows that
c=2,b=3c−3=6−3=3,a=1−b−2c=1−3−4=−6.
We conclude that (1,−2,5)=−6(1,1,1)+3(1,2,3)+2(2,−1,1).
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