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Math Help - Linear Algebra

  1. #1
    Member
    Joined
    Mar 2006
    Posts
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    Linear Algebra

    Hello,

    I am doing my math homework, and I come across two problems I can't tend to do. Something always come up wrong. If anyone can help me, it is truly appreciated. Thank you so much.

    I have to find the reduce echelon form for these two problems

    <br />
A=\[ \left( \begin{array}{ccccc}1 & -2 & 1 & 1 & 4\\2 & 1 & -3 & -1 & 6\\1 & -7 & -6 & 2 & 6\end{array} \right)\]<br />


    <br />
A=\[ \left( \begin{array}{ccccc}0 & 1 & 2 & 1\\2 & 1 & 0 & 2\\0 & 2 & 1 & 1\end{array} \right)\]<br />
    Last edited by jenjen; May 2nd 2006 at 07:17 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jenjen
    Hello,

    I am doing my math homework, and I come across two problems I can't tend to do. Something always come up wrong. If anyone can help me, it is truly appreciated. Thank you so much.

    I have to find the reduce echelon form for these two problems

    <br />
A=\[ \left( \begin{array}{ccccc}1 & -2 & 1 & 1 & 4\\2 & 1 & -3 & -1 & 6\\1 & -7 & -6 & 2 & 6\end{array} \right)\]<br />
    <br />
A=\[ \left( \begin{array}{ccccc}<br />
1 & -2 & 1 & 1 & 4\\<br />
2 & 1 & -3 & -1 & 6\\<br />
1 & -7 & -6 & 2 & 6<br />
\end{array} \right)\]<br />

    Subtract twice the first row from the second, and subtract the first row from
    the third:

    <br />
B=\[ \left( \begin{array}{ccccc}<br />
1 & -2 & 1 & 1 & 4\\<br />
2-2 & 1+4 & -3-2 & -1-2 & 6-8\\<br />
1-1 & -7+2 & -6-1 & 2-1 & 6-4<br />
\end{array} \right)\]<br />
<br />
=\[ \left( \begin{array}{ccccc}<br />
1 & -2 & 1 & 1 & 4\\<br />
0 & 5 & -5 & -3 & -2\\<br />
0 & -5 & -7 & 1 & 2<br />
\end{array} \right)\]<br />

    Now add the second row to the last row:

    <br />
C=\[ \left( \begin{array}{ccccc}<br />
1 & -2 & 1 & 1 & 4\\<br />
0 & 5 & -5 & -3 & -2\\<br />
0 & -5+5 & -7-5 & 1-3 & 2-2<br />
\end{array} \right)\]<br />
] <br />
=\[ \left( \begin{array}{ccccc}<br />
1 & -2 & 1 & 1 & 4\\<br />
0 & 5 & -5 & -3 & -2\\<br />
0 & 0 & -12 & -2 & 0<br />
\end{array} \right)\]<br />

    Now finally divide the second row through by 5, and the third row by -12 to get:

    <br />
\[ \left( \begin{array}{ccccc}<br />
1 & -2 & 1 & 1 & 4\\<br />
0 & 1 & -1 & -3/5 & -2/5\\<br />
0 & 0 & 1 & 1/6 & 0<br />
\end{array} \right)\]<br />

    RonL
    Last edited by CaptainBlack; May 5th 2006 at 07:39 AM.
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