# Thread:

1. ## 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: $3t+1 \not= \frac{1}{3}-\frac{2}{3}(3t^2+1)$

2. $P_1=2t^2+t+1,P_2=3t+1,P_3=3t^2+1$.
$P_1=\frac{1}{3}P_2+\frac{2}{3}P_3$.

3. $2t^2+t+1=a(3t+1)+b(3t^2+1)\Rightarrow \left\{\begin{matrix}<br /> 3b=2 \\ <br /> 3a=1 \\ <br /> a+b=1 <br /> \end{matrix}\right.$ $\Rightarrow a=\frac{1}{3},b=\frac{2}{3}$
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