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Math Help - Finding a basis of R^3

  1. #1
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    Finding a basis of R^3

    Find a basis of R^3 defined by the equation,

    2x_{1}+3x_{2}+x_{3} = 0

    I took the above equation and solved for x3, assuming that was saying that x3 could be created using -2 copies of x1 minus 3 copies of x2...

    2x_{1}+3x_{2} = -x_{3}
    -2x_{1}-3x_{2} = x_{3}

    Then I solved this like I solve for a kernel.

    \begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}=\begin{bmatrix}t\\ s\\ -2t-3s\end{bmatrix}= t\begin{bmatrix}1\\ 0\\ -2\end{bmatrix}+s\begin{bmatrix}0\\ 1\\ -3\end{bmatrix}

    Which gave me a result of,

    \begin{bmatrix}1\\ 0\\ -2\end{bmatrix},\begin{bmatrix}0\\ 1\\ -3\end{bmatrix}

    However, the answer is:

    \begin{bmatrix}-3\\ 2\\ 0\end{bmatrix},\begin{bmatrix}-1\\ 0\\ 2\end{bmatrix}

    Where did I go wrong? Any help is extremely appreciated. Thanks!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by tangibleLime View Post
    Where did I go wrong? Any help is extremely appreciated. Thanks!

    You are completely right. Take into account that the solution is not unique.
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  3. #3
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    Ah, thank you for the clarification!
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  4. #4
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    Also note that the problem should be "Find a basis of the subspace of R^3 defined by the equation,
    2x_1+ 3x_2+ x_3= 0"

    Since that is one equation constraining the three variables, the subspace has dimension 3- 1= 2 and so your basis should contain 2 vectors, just as you say.
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