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Math Help - Find a set of vectors spanning the solution space, stuck at last step

  1. #1
    Member Jskid's Avatar
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    Find a set of vectors spanning the solution space, stuck at last step

    Find a set of vectors spanning the solution space of Ax=0 where <br />
\[<br />
A =<br />
\left[ {\begin{array}{cccc}<br />
 1 & 0 & 1 & 0 \\<br />
 1 & 2 & 3 & 1 \\<br />
 2 & 1 & 3 & 1 \\<br />
 1 & 1 & 2 & 1 \\<br />
 \end{array} } \right]<br />
\]<br />
    So I rref and it's row equivalent to the identity matrix. How do I interpret the result?
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  2. #2
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    Quote Originally Posted by Jskid View Post
    Find a set of vectors spanning the solution space of Ax=0 where <br />
\[<br />
A =<br />
\left[ {\begin{array}{cccc}<br />
 1 & 0 & 1 & 0 \\<br />
 1 & 2 & 3 & 1 \\<br />
 2 & 1 & 3 & 1 \\<br />
 1 & 1 & 2 & 1 \\<br />
 \end{array} } \right]<br />
\]<br />
    So I rref and it's row equivalent to the identity matrix. How do I interpret the result?


    If the matrix is row-equivalent to the unit matrix then the only solution to the adjoint homogeneous

    system is the trivial one...

    Tonio
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  3. #3
    Member Jskid's Avatar
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    I did it wrong, rref(A) = \[<br />
\left[ {\begin{array}{cccc}<br />
 1 & 0 & 1 & 0  \\<br />
 0 & 1 & 1 & 0  \\<br />
 0 & 0 & 0 & 1  \\<br />
 0 & 0 & 0 & 0  \\<br />
 \end{array} } \right]<br />
\]<br />
    So x_1=-r
    x_2=-r
    x_3=r
    x_4 I'm not sure what to do with
    Last edited by Jskid; February 27th 2011 at 03:06 PM. Reason: minor
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  4. #4
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    Have you considered
    \[<br />
\left[ {\begin{array}{r}   1  \\   1  \\   { - 1}  \\   0  \\ \end{array} } \right]~?
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  5. #5
    Member Jskid's Avatar
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    I didn't realize a row of 0s could be -r I thought it had to be r
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  6. #6
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    Quote Originally Posted by Jskid View Post
    I did it wrong, rref(A) = \[<br />
\left[ {\begin{array}{cccc}<br />
 1 & 0 & 1 & 0  \\<br />
 0 & 1 & 1 & 0  \\<br />
 0 & 0 & 0 & 1  \\<br />
 0 & 0 & 0 & 0  \\<br />
 \end{array} } \right]<br />
\]<br />
    So x_1=-r
    x_2=-r
    x_3=r
    x_4 I'm not sure what to do with
    I have no clue what "r" means here. Is it just an arbitrary number?

    The equations corresponding to your reduced matrix are x_1+ x_3= 0, x_2+ x_3= 0, and x_4= 0. The first two equations can be solved for x_1= -x_3, x_2= -x_3. Taking x_3 to be your "r", we have x_1= -r, x_2= -r, x_3= r and, of course, x_4= 0. Taking r= -1 gives Plato's solution and all solutions are a multiple of that.
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