Is this the correct method for finding a coordinate vector?

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SUMMARY

The discussion confirms that the method of using Gaussian elimination to find a coordinate vector is correct. The user solved the system of equations: \(\frac{1}{2} = x + \frac{1}{2}y + \frac{1}{3}z\), \(\frac{1}{3} = -x + \frac{1}{2}y\), and \(-\frac{9}{2} = x + z\), resulting in the values \(x=1\), \(y=\frac{8}{3}\), and \(z=-\frac{11}{2}\). The final coordinate vector is confirmed as \(\begin{bmatrix} 1 \\ \frac{8}{3} \\ -\frac{11}{2} \end{bmatrix}\). This method is validated by the participant Ted123.

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Homework Statement



[PLAIN]http://img25.imageshack.us/img25/3409/linearf.jpg

Homework Equations



The Attempt at a Solution



Is this the right method to do this type of question (as I haven't seen an example of finding a coordinate vector before)?

Do I solve the following system of equations?

\displaystyle \frac{1}{2} = x + \frac{1}{2}y + \frac{1}{3}z

\displaystyle \frac{1}{3} = -x + \frac{1}{2}y

\displaystyle -\frac{9}{2} = x+z

If so by Guassian Elimination I get

x=1,\,y=\frac{8}{3},\,z=-\frac{11}{2}

So is \begin{bmatrix} 1 \\ 8/3 \\ -11/2 \\ \end{bmatrix} the coordinate vector?
 
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Hi Ted123! :wink:

Yes, that's fine! :biggrin:
 

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