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Math Help - Linearly independent polynomials; answer not checking out.

  1. #1
    Member Jskid's Avatar
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    Linearly independent polynomials; answer not checking out.

    Are the following vectors in P_2 linearly dependent? If yes, express one vector as a linear combination of the rest.
    {3t+1,3t^2+1,2t^2+t+1}

    This forms the matrix <br />
\[<br /> <br />
\left[ {\begin{array}{ccc}<br />
 0 & 3 & 2 \\<br />
 3 & 0 & 1  \\<br />
 1 & 1 & 1 \\<br />
 \end{array} } \right]<br />
\] which is row equivalent to \[<br /> <br />
\left[ {\begin{array}{ccc}<br />
 1 & 0 & \frac{1}{3}  \\<br />
 0 & 1 & \frac{2}{3}  \\<br />
 0 & 0 & 0 \\<br />
 \end{array} } \right]<br />
\]
    So c_1=\frac{r}{3},c_2=\frac{2r}{3}, c_3=-r is a solution for the homogeneous system, where r is any real number.
    Let r = 1, then \frac{1}{3}P_1 + \frac{2}{3}P_2-1=0
    Solving for P_1=\frac{1}{3}-\frac{2}{3}P_2
    But when I check it doesn't work: 3t+1 \not= \frac{1}{3}-\frac{2}{3}(3t^2+1)
    Last edited by Ackbeet; March 1st 2011 at 02:26 AM. Reason: Fixed spelling and grammar in title.
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  2. #2
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    P_1=2t^2+t+1,P_2=3t+1,P_3=3t^2+1.
    P_1=\frac{1}{3}P_2+\frac{2}{3}P_3.
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  3. #3
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    2t^2+t+1=a(3t+1)+b(3t^2+1)\Rightarrow \left\{\begin{matrix}<br />
3b=2  \\ <br />
3a=1  \\ <br />
a+b=1 <br />
\end{matrix}\right. \Rightarrow a=\frac{1}{3},b=\frac{2}{3}
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