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Thread: Nullspace

  1. #1
    Rui
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    Nullspace

    Ax=b
    a) calculate the nullspace Null(A) and
    b) give the general solution of the system Ax=b or state that the system inconsistent.
    For b) write in the form x= p+y where p is a particular solution and y ranges across Null (A)

    A=1 1 0 2
    2 1 1 -2
    2 2 2 1

    b= 12
    0
    14
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Rui View Post
    Ax=b
    a) calculate the nullspace Null(A) and
    b) give the general solution of the system Ax=b or state that the system inconsistent.
    For b) write in the form x= p+y where p is a particular solution and y ranges across Null (A)

    A=1 1 0 2
    2 1 1 -2
    2 2 2 1

    b= 12
    0
    14
    To find the Null space solve the matrix equation Ax=0

    $\displaystyle \begin{bmatrix}
    1 && 1 && 0 && 2 && 0 \\
    2 && 1 && 1 && -2 && 0\\
    2 && 2 && 2 && 1 && 0 \\
    \end{bmatrix} = \begin{bmatrix}
    1 && 0 && 0 && -5/2 && 0 \\
    0 && 1 && 0 && 9/2 && 0\\
    0 && 0 && 1 && -3/2 && 0 \\
    \end{bmatrix}$

    now is $\displaystyle x_4=t$ then $\displaystyle x_3=3t/2$
    $\displaystyle x_2=-9t/2$ $\displaystyle x_1=5t/2 $

    so the vector that spans the Null Space is

    $\displaystyle v= \begin{bmatrix}
    5t/2 \\
    -9t/2 \\
    3t/2 \\
    t \\
    \end{bmatrix} = t/2 \begin{bmatrix}
    5 \\
    -9 \\
    3 \\
    2 \\

    \end{bmatrix} = s \begin{bmatrix}
    5 \\
    -9 \\
    3 \\
    2 \\

    \end{bmatrix}

    $

    solving the system Ax=b

    $\displaystyle \begin{bmatrix}
    1 && 1 && 0 && 12 && 0 \\
    2 && 1 && 1 && 0 && 0\\
    2 && 2 && 2 && 14 && 0 \\
    \end{bmatrix} = \begin{bmatrix}
    1 && 0 && 0 && -5/2 && -7 \\
    0 && 1 && 0 && 9/2 && 19\\
    0 && 0 && 1 && -3/2 && -5 \\
    \end{bmatrix}$

    A particualr solution is
    $\displaystyle \begin{bmatrix}
    -7 \\
    19 \\
    -5 \\
    0 \\
    \end{bmatrix}$

    so finally we get all solutions are of the form

    $\displaystyle \begin{bmatrix}
    x_1 \\
    x_2 \\
    x_3 \\
    x_4 \\
    \end{bmatrix}= s \begin{bmatrix}
    5 \\
    -9 \\
    3 \\
    2 \\

    \end{bmatrix}+ \begin{bmatrix}
    -7 \\
    19 \\
    -5 \\
    0 \\
    \end{bmatrix}
    $
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