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Linear Combination of Vectors

  1. Apr 8, 2005 #1
    The question asks: Write down the vector b as the linear combination of vectors, v1, v2, v3.

    To which I got:

    b = [tex]\left(\begin{array}{cc}1\\3\\2\end{array}\right)[/tex] = [tex]\left(\begin{array}{cc}1\\-2\\2\end{array}\right)\frac{7}{2} + \left(\begin{array}{cc}0\\1\\-3\end{array}\right)0 + \left(\begin{array}{cc}-1\\4\\-2\end{array}\right)\frac{5}{2}[/tex]

    (as A was the matrix made up of those three vectors, and the scalars were the answers from the system Ax=b)

    Then the question asks: Determine whether the vector v1 is the linear combination of vectors, v2, v3, b.

    Could someone tell me how to find the scalar for vector b?

    At the moment I have:

    [tex]\left(\begin{array}{cc}0\\1\\-3\end{array}\right)0+ \left(\begin{array}{cc}-1\\4\\-2\end{array}\right)\frac{5}{2}+ \left(\begin{array}{cc}\frac{7}{2}\\0\\\frac{5}{2}\end{array}\right)[/tex]

    And I don't know what number to multiply the last vector by...
     
  2. jcsd
  3. Apr 8, 2005 #2
    Shouldnt the linear combinations look like this?

    [tex]
    \left(\begin{array}{cc}1\\3\\2\end{array}\right)=
    1\left(\begin{array}{cc}1\\0\\0\end{array}\right)+
    3\left(\begin{array}{cc}0\\1\\0\end{array}\right)+
    2\left(\begin{array}{cc}0\\0\\1\end{array}\right)
    [/tex]
     
  4. Apr 8, 2005 #3

    arildno

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    No, he were to express the vector b as a linear combination in terms of the three originally given vectors.
     
  5. Apr 8, 2005 #4

    arildno

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    shan:
    Just multiply your original (vector) equation with [tex]\frac{2}{7}[/tex] and rearrange your terms.
     
  6. Apr 8, 2005 #5

    dextercioby

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    Erenion,u used 3 vectors that could form a basis in [itex]\mathbb{C}^{3} [/itex]...I think this property will be lost,once u try to express one of the basis vectors as a lin.combo.of the other basis vectors & the original vector...

    Since the OP's problem dooesn't mention linear independence & completitude,i guess what he's done is correct and he can solve point "b",if he's more careful with his algebra...

    Daniel.
     
  7. Apr 8, 2005 #6
    hmm but I don't understand why? As I need to determine whether v1 is a linear combination of vectors v2, v3 and b, if I multiplied the original vector equation with [tex]\frac{2}{7}[/tex], wouldn't I change the length of v1?
     
  8. Apr 9, 2005 #7

    arildno

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    What do you mean?
    v1 is the vector (1,-2,2), right?
     
  9. Apr 9, 2005 #8
    Yeah... ok maybe I'm misunderstanding. By original vector equation, you mean this one?

    [tex]\left(\begin{array}{cc}1\\-2\\2\end{array}\right)\frac{7}{2} + \left(\begin{array}{cc}0\\1\\-3\end{array}\right)0 + \left(\begin{array}{cc}-1\\4\\-2\end{array}\right)\frac{5}{2}[/tex]

    So you're saying to multiply the above equation by \frac{2}{7} and rearrange to make v1 (1, -2, 2) the subject?
     
  10. Apr 9, 2005 #9

    arildno

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    I take it you meant:
    b=[tex]\left(\begin{array}{cc}1\\-2\\2\end{array}\right)\frac{7}{2} + \left(\begin{array}{cc}0\\1\\-3\end{array}\right)0 + \left(\begin{array}{cc}-1\\4\\-2\end{array}\right)\frac{5}{2}[/tex]

    Yes, that's the equation I was referring to..
     
  11. Apr 9, 2005 #10
    Ah ok, I got it. Thanks for the help :)
     
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