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Math Help - Nullspace Q

  1. #1
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    Nullspace Q

    The question reads: "For the matrix A below, find a spanning set for its Nullspace.
    A=\left( \begin{array}{cccc} 1 & 3 & -2 & 1 \\ 2 & 1 & 3 & 2 \\ 3 & 4 & 5 & 6 \end{array}\right)"


    So... I want to solve for x such that Ax = 0.

    I row reduce; r2 -> r2-2r1, r3->r3-3r1

    A=\left( \begin{array}{cccc} 1 & 3 & -2 & 1 \\ 0 & -5 & 7 & 0 \\ 0 & -5 & 11 & 3 \end{array}\right)

    then r3->r3-r2

    A=\left( \begin{array}{cccc} 1 & 3 & -2 & 1 \\ 0 & -5 & 7 & 0 \\ 0 & 0 & 4 & 3 \end{array}\right)

    Giving x_{1}+3x_{2}-2x_{3}+x_{4} = 0
    -5x_{2}+7x_{3} = 0<br />
4x_{3}+3x_{4}.

    So we can say x_{4} = \frac{4x_{3}}{3}, x_{3}=x_{3}, x_{2} = \frac {7x_{3}}{5} , x_{1} = \frac{-26x_{3}}{15}.

    So the Nullspace is spanned by \left (\begin{array}{c} \frac{-26}{15} \\ \frac{7}{15} \\ 1 \\ \frac{4}{3} \end{array}\right)x_{3}

    Should Ax not then be zero?

    \left(\begin{array}{cccc} 1 & 3 & -2 & 1 \\ 0 & -5 & 7 & 0 \\ 0 & -5 & 11 & 3 \end{array}\right) \left (\begin{array}{c} \frac{-26}{15} \\ \frac{7}{15} \\ 1 \\ \frac{4}{3} \end{array}\right)x_{3} = \left(\begin{array}{c} -1 \\ \frac{8}{3} \\ \frac{29}{3} \end{array} \right)

    I'm obviously wrong somewhere... but I can't spot where...
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  2. #2
    Senior Member
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    Thanks
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    starting off...

    Your last equation has
    <br />
4 x_3 + 3 x_4 = 0<br />
    so shouldn't
    <br />
x_4 = - 4 x_3 / 3<br />
    Note the negative sign, and keep on going...
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  3. #3
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    Got it. Thank you.
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