1. ## Linearly independent polynomials; answer not checking out.

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

This forms the matrix $\displaystyle $\left[ {\begin{array}{ccc} 0 & 3 & 2 \\ 3 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array} } \right]$$ which is row equivalent to $\displaystyle $\left[ {\begin{array}{ccc} 1 & 0 & \frac{1}{3} \\ 0 & 1 & \frac{2}{3} \\ 0 & 0 & 0 \\ \end{array} } \right]$$
So $\displaystyle c_1=\frac{r}{3},c_2=\frac{2r}{3}, c_3=-r$ is a solution for the homogeneous system, where r is any real number.
Let r = 1, then $\displaystyle \frac{1}{3}P_1 + \frac{2}{3}P_2-1=0$
Solving for $\displaystyle P_1=\frac{1}{3}-\frac{2}{3}P_2$
But when I check it doesn't work: $\displaystyle 3t+1 \not= \frac{1}{3}-\frac{2}{3}(3t^2+1)$

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

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