if a1=(7,-1,4) a2=(4,0,1) and a3=(-3,5,0) and the set S={a1,a2,a3} is a basis for R3, how do you find the equation y=2a1-3a2+a3 (ps that is the answer, I just don't know how to get to it)

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if a1=(7,-1,4) a2=(4,0,1) and a3=(-3,5,0) and the set S={a1,a2,a3} is a basis for R3, how do you find the equation y=2a1-3a2+a3 (ps that is the answer, I just don't know how to get to it)

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trying to find the linear combination y, they give the answer but they dont explain how they got to it
perhaps I am reading this wrong
do you know the coordinates of y?

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y=(-1,3,5)
didnt think it was part of it, but they did mention that a little before
yes, that is the important bit. a1(x,y,z) = (7,-1,4) a2(x,y,z) = (4,0,1) and a3(x,y,z) = (-3,5,0)
right, so how do we find the linear combination?
so a linear combination of the x-coordinates of a1,a2,a3 should give the x- coordinate of y, a linear combination of the y-coordinates of a1,a2,a3 should give the y- coordinate of y, a linear combination of the z-coordinates of a1,a2,a3 should give the z- coordinate of y.
so 7m+4n-3p=-1 -m+5p=3 and4m+n = 5
ahh ok thanks a lot, then solve the matrix from there eh?
3 equations, 3 unknowns, solve to get m,n,p those will be the coefficients of the equation y = ma1+na2+pa3
just one more thing....what did we do here? we had a coordinate, then we established the basis, and found another coordinate
did we change it from cartesian?
no its in Cartesian coordinates only. all we did was to find a linear combination of the three points which gives the point y.
it follows that [y]s=(2,-3,1) in the text, this isn't a coordinate?
no, those are the coefficients of the linear combination.
i gotta run, hope that helped!
thanks!

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