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Math Help - [SOLVED] Linear combination of a vector S

  1. #1
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230

    [SOLVED] Linear combination of a vector S

    Let S be = <br />
\begin{bmatrix}2 \\ 1 \\ 0 \\ 4 \\ \end{bmatrix} \begin{bmatrix}1 \\ 3 \\ 4 \\ -4 \\ \end{bmatrix} \begin{bmatrix}3 \\ 4 \\ 4 \\ 0 \\ \end{bmatrix} \begin{bmatrix}4 \\ -3 \\ -3 \\ -2 \\ \end{bmatrix} <br />

    and let W be the subspace spanned by S.

    U{\rightharpoonup} = \begin{bmatrix}9 \\ 2 \\ 1 \\ 2 \\ \end{bmatrix}

    Is U{\rightharpoonup} in W? If so find a linear combination of the vectors in S Which adds up to U

    So i know that i am trying to find r1, r2, r3, r4, such that

    <br />
r1*\begin{bmatrix}2 \\ 1 \\ 0 \\ 4 \\ \end{bmatrix} + r2*\begin{bmatrix}1 \\ 3 \\ 4 \\ -4 \\ \end{bmatrix} + r3*\begin{bmatrix}3 \\ 4 \\ 4 \\ 0 \\ \end{bmatrix} +r4*\begin{bmatrix}4 \\ -3 \\ -3 \\ -2 \\ \end{bmatrix} = \begin{bmatrix}9 \\ 2 \\ 1 \\ 2 \\ \end{bmatrix}<br />

    So i found the reduced row echelon form of the matrix and this is what i got

    \begin{bmatrix}1&0&1&0&|2 \\ 0&1&1&0&|1 \\ 0&0&0&1&|1 \\ 0&0&0&0&|0 \\ \end{bmatrix}<br />

    column 1 is x1 column 2 is x2 .... etc.

    From this i can see that the 3rd column is free which shows that there are infinitely many solutions. So YES U is in W because the system has at least one solution.

    so i paramaterized the solutions and set....

    x3 = y
    x1 + x3 = 2
    x2 + x3 = 1
    x4 = 1

    so

    x1 = 2-y
    x2 = 1-y
    x3 = y
    x4 = 1

    y can be any real # so... if y=1 then

    r1 = 1
    r2 = 0
    r3 = 1
    r4 = 1

    This ^^ is one linear combination of the vectors in S that adds up to U

    Did i do all this right? I just want to check. Thanks for looking it over and commenting on anything
    Last edited by mybrohshi5; March 23rd 2010 at 09:25 PM.
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  2. #2
    Member mathemagister's Avatar
    Joined
    Feb 2010
    Posts
    191
    Quote Originally Posted by mybrohshi5 View Post
    Let S be = <br />
\begin{bmatrix}2 \\ 1 \\ 0 \\ 4 \\ \end{bmatrix} \begin{bmatrix}1 \\ 3 \\ 4 \\ -4 \\ \end{bmatrix} \begin{bmatrix}3 \\ 4 \\ 4 \\ 0 \\ \end{bmatrix} \begin{bmatrix}4 \\ -3 \\ -3 \\ -2 \\ \end{bmatrix} <br />

    and let W be the subspace spanned by S.

    U{\rightharpoonup} = \begin{bmatrix}9 \\ 2 \\ 1 \\ 2 \\ \end{bmatrix}

    Is U{\rightharpoonup} in W? If so find a linear combination of the vectors in S Which adds up to U

    So i know that i am trying to find r1, r2, r3, r4, such that

    <br />
r1*\begin{bmatrix}2 \\ 1 \\ 0 \\ 4 \\ \end{bmatrix} + r2*\begin{bmatrix}1 \\ 3 \\ 4 \\ -4 \\ \end{bmatrix} + r3*\begin{bmatrix}3 \\ 4 \\ 4 \\ 0 \\ \end{bmatrix} +r4*\begin{bmatrix}4 \\ -3 \\ -3 \\ -2 \\ \end{bmatrix} = \begin{bmatrix}9 \\ 2 \\ 1 \\ 2 \\ \end{bmatrix}<br />

    So i found the reduced row echelon form of the matrix and this is what i got

    \begin{bmatrix}1&0&1&0&|2 \\ 0&1&1&0&|1 \\ 0&0&0&1&|1 \\ 0&0&0&0&|0 \\ \end{bmatrix}<br />

    column 1 is x1 column 2 is x2 .... etc.

    From this i can see that the 3rd column is free which shows that there are infinitely many solutions. So YES U is in W because the system has at least one solution.

    so i paramaterized the solutions and set....

    x3 = y
    x1 + x3 = 2
    x2 + x3 = 1
    x4 = 1

    so

    x1 = 2-y
    x2 = 1-y
    x3 = y
    x4 = 1

    y can be any real # so... if y=1 then

    r1 = 1
    r2 = 0
    r3 = 1
    r4 = 1

    This ^^ is one linear combination of the vectors in S that adds up to U

    Did i do all this right? I just want to check. Thanks for looking it over and commenting on anything
    Although I am not extremely knowledgeable in this field, it looks like you did everything correctly. I looked at each of your steps and they all look correct.

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