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Linear combination

  1. Jun 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]V=\mathbb(R)^3[/tex]. Find all values of k for which the vector u is a linear combination of the vectors given below:

    [tex]v_1=(2,3,5)[/tex]; [tex]v_2=(3,7,8)[/tex]; [tex]v_3=(1,-6,1)[/tex] y [tex]u=(7,-2,k)[/tex]

    [tex]\begin{Bmatrix}{ 2\lambda_1+3\lambda_2+\lambda_3=7} \\3\lambda_1+7\lambda_2-6\lambda_3=-2 \\5\lambda_1+8\lambda_2+\lambda_3=k \end{matrix}[/tex]

    [tex]\begin{bmatrix}{2}&{3}&{1}&{7}\\{3}&{7}&{-6}&{-2}\\{5}&{8}&{3}&{k}\end{bmatrix}\rightarrow{
    \begin{bmatrix}{2}&{3}&{1}&{7}\\{0}&{5/2}&{-15/2}&{-25/2}\\{0}&{1/2}&{1/2}&{k-35/2}\end{bmatrix}}\rightarrow{
    \begin{bmatrix}{2}&{3}&{1}&{7}\\{0}&{5}&{-15}&{-25}\\{0}&{0}&{32}&{k-35}\end{bmatrix}}[/tex]


    [tex]\begin{bmatrix}{2}&{3}&{1}&{7}\\{0}&{1}&{-3}&{-5}\\{0}&{0}&{3}&{k-30}\end{bmatrix}\rightarrow{\begin{bmatrix}{2}&{3}&{0}&{\displaystyle\frac{-k}{3}+10}\\{0}&{1}&{0}&{k-35}\\{0}&{0}&{3}&{k-30}\end{bmatrix}}\rightarrow{\begin{bmatrix}{2}&{0}&{0}&{\displaystyle\frac{-10}{3}k+115}\\{0}&{1}&{0}&{k-35}\\{0}&{0}&{1}&{k/3-10}\end{bmatrix}}\rightarrow{\begin{bmatrix}{1}&{0}&{0}&{\displaystyle\frac{-5}{3}k+115/2}\\{0}&{1}&{0}&{k-35}\\{0}&{0}&{1}&{k/3-10}\end{bmatrix}}[/tex]

    [tex]\begin{Bmatrix}{ \lambda_1=\displaystyle\frac{-5}{3}k+\displaystyle\frac{115}{2}} \\ \lambda_2=k-35 \\ \lambda_3=\displaystyle\frac{k}{3}-10 \end{matrix}[/tex]

    Is that right?
     
  2. jcsd
  3. Jun 14, 2010 #2

    vela

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    You seem to have made a mistake in the second step.
     
  4. Jun 14, 2010 #3
    Oh shhhhhhh.... ok. Anyway. What does it mean when I got this kind of solution in terms of vectorial spaces? I mean, when its undetermined. And if I get an incompatible solution, what would it mean?

    Bye and thanks.
     
  5. Jun 14, 2010 #4

    Mark44

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    Your matrix is really an augmented matrix, with the first three columns being the three given vectors, and the fourth column being the vector u. If you end up with a row whose first three entries are 0 and the fourth is an some expression involving k, the system is inconsistent if the expression involving k is nonzero. The system is consistent if and only if the expression involving k is equal to zero.
     
  6. Jun 14, 2010 #5

    vela

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    And if your initial work had been correct, that would mean for any value of k you could find a linear combination that would work, so your answer should have been "all real numbers."
     
  7. Jun 15, 2010 #6
    Thanks Mark and vela.
     
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