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Math Help - basis of null

  1. #1
    Newbie
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    May 2009
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    basis of null

    I attached the problem.

    I'm totally confusing..

    Can anyone help?
    Attached Thumbnails Attached Thumbnails basis of null-aa.jpg  
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  2. #2
    Junior Member
    Joined
    Apr 2009
    Posts
    58
    First you should row reduce the matrix M into echelon form:
    you should get:
    <br />
B = \begin{bmatrix}1&2&1&3&0&0\\0&1&-1&2&-2&3\\0&0&0&1&1&-3\\0&0&0&0&1&-2\end{bmatrix}<br />

    For the basis of the column of M, you look at the pivot columns of B, these are the first, second, fourth, and fifth columns. Then you use these columns from M as the basis of column, so the answer for that should be:

    <br />
  Col M= \begin{bmatrix}1&2&-1&0\end{bmatrix}, \begin{bmatrix}2&5&-1&0\end{bmatrix}, <br />
yeahh those are meant to be written as vectors and don't forget the fourth and fifth columns

    To find the basis of the Null M, you just have to solve  Mx= 0
    yeahhh and express the set via free variables, (you should have 2 free variables so the answer should be expressed using  x_3 and  x_5.

    To find the basis of the Row of M, use the row echelon form. So the answer should be:
    <br />
Row M = \begin{bmatrix}1&2&1&3&0&0\end{bmatrix}, <br />
\begin{bmatrix}0&1&-1&2&-2&3\end{bmatrix}, <br />
\begin{bmatrix}0&0&0&1&1&-3\end{bmatrix},<br />
\begin{bmatrix}0&0&0&0&1&-2\end{bmatrix}<br />

    yupp that's it
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