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Math Help - matrix row reduction solution question..

  1. #1
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    matrix row reduction solution question..

    \begin{pmatrix}<br />
1 &  1& \lambda  &| 1\\ <br />
 0&  \lambda -1  &1-\lambda   &| 0\\ <br />
 0&1  &1-\lambda      &|1-\lambda ^2 <br />
\end{pmatrix}

    cant we say that for lambda=1 we have endless solutions?

    but also we can divide equation 2 and 3 by lambda-1 and after that so for lambda=1 we get that there is no solutions
    so what is the correct solution
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  2. #2
    Member HappyJoe's Avatar
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    You know, when \lambda=1, we have that \lambda-1=0. So in this case, it doesn't make sense to divide by \lambda-1.
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  3. #3
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    Quote Originally Posted by transgalactic View Post
    \begin{pmatrix}<br />
1 &  1& \lambda  &| 1\\ <br />
 0&  \lambda -1  &1-\lambda   &| 0\\ <br />
 0&1  &1-\lambda      &|1-\lambda ^2 <br />
\end{pmatrix}

    cant we say that for lambda=1 we have endless solutions?

    but also we can divide equation 2 and 3 by lambda-1 and after that so for lambda=1 we get that there is no solutions
    so what is the correct solution
    With \lambda= 1 the matrix is
    \begin{pmatrix}<br />
1 &  1& 1  &| 1\\ <br />
 0& 0  & 0   &| 0\\ <br />
 0&1  & 0      &| 0\end{pmatrix}

    Which corresponds to the equations x+ y+ z= 1, 0= 0, 0= 0. It should be obvious that you can choose, say, y and z to be any numbers you want and solve for the corresponding x. That is why there are an infinite number of solutions.

    As HappyJoe said (may he long be happy), you can only divide by \lambda- 1 as long as \lambda- 1\ne 0 so that it is not valid to do that and then let \lambda= 1.
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