Solving Linear Combinations: (1,2,3)

Precursor
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Homework Statement
Write the vector (1,2,3) as a linear combination of the vectors (1,0,1), (1,0,-1), and (0,1,1).

The attempt at a solution
(1,2,3) = C1(1,0,1) + C2(1,0,-1) + C3(0,1,1)

The matrix for this is:

1...1...0...1
0...0...1...2
1...-1...1...3

I reduced it to the following:

1...0...0...1
0...1...0...0
0...0...1...2

Therefore, (1,2,3) is a linear combination of the vectors (1,0,1), (1,0,-1), and (0,1,1). Am I right?
 
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Easy check:

\mbox{Does }\left( \begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right) = \textbf{1}\, \left( \begin{array}{c}1 \\ 0 \\ 1 \end{array}\right) + \, \textbf{0}\, \left( \begin{array}{c}1 \\ 0 \\ -1 \end{array}\right) + \, \textbf{2}\, \left( \begin{array}{c}0 \\ 1 \\ 1 \end{array}\right) \, ?​


You made an arithmedic error reducing the augmented matrix, try again...
 
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Do you even need that middle vector...?
 
benorin said:
You made an arithmetic error reducing the augmented matrix, try again...
Looks right to me.
 
Precursor said:
Homework Statement
Write the vector (1,2,3) as a linear combination of the vectors (1,0,1), (1,0,-1), and (0,1,1).

...

Therefore, (1,2,3) is a linear combination of the vectors (1,0,1), (1,0,-1), and (0,1,1). Am I right?
Yes, it's a linear combination of those vectors, but you should explicitly write out what that linear combination is because that's what the problem asked for.
 
Ok, so all I need to do is substitute in C1, C2, and C3 in front of the appropriate vectors in the original equation?
 
Yes. And as benorin mentioned, it's very easy to check.
 
One idea conveyed here is that one may use any linearly independent set of vectors to describe a space. Cartesian (sic?) coordinates use the standard basis vectors so that the (x,y,z) style coordinate (1,2,3) is a linear combination of the vectors (1,0,0), (0,1,0), and (0,0,1). Namely,

(1,2,3) = 1*(1,0,0) + 2*(0,1,0) + 3* (0,0,1)​

But, other than their linear independence, these are not special. If you have studied linear independence, deter if the 3 vectors used in the problem match this requirement. They needn't even be boring, stick-arrow vectors, polar, cylindrical, spherical coordinates also work.
 
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