The set $\displaystyle {1+t,t-t^2,1+2t-t^2}$ is a base of P2?Why?
I think you mean a basis...
First off, in order for this set to be a basis of $\displaystyle P_2$, it must have span $\displaystyle P_2$
This means that $\displaystyle a_1 (1+t)+a_2(t-t^2)+a_3(1+2t-t^2)=b_1t^2+b_2t+b_3$ $\displaystyle \implies (-a_2-a_3)t^2+(a_1+a_2+3a_3)t+a_1+a_3=b_1t^2+b_2t+b_3$
We see that the corresponding augmented matrix is $\displaystyle \left[\begin{array}{ccc|c}0&-1&-1&b_1\\1&1&1&b_2\\1&0&1&b_3\end{array}\right]$
Getting it into RREF (I leave this for you to do), we see that $\displaystyle \left[\begin{array}{ccc|c}1&0&0&b_1+b_2\\0&1&0&b_2-b_3\\0&0&1&-b_1-b_2+b_3\end{array}\right]$
Since the system has solutions (primarily $\displaystyle a_1=b_1+b_2,~a_2=b_2-b_3,$ and $\displaystyle a_3=-b_1-b_2+b_3$), we see that $\displaystyle \left\{1+t,t-t^2,1+2t-t^2\right\}$ spans $\displaystyle P_2$.
Now, we need to show that these elements are linearly independent.
$\displaystyle a_1 (1+t)+a_2(t-t^2)+a_3(1+2t-t^2)=0$
This can be rewritten as $\displaystyle (-a_2-a_3)t^2+(a_1+a_2+3a_3)t+a_1+a_3=0$
Thus, an augmented matrix can be constructed:
$\displaystyle \left[\begin{array}{ccc|c}0&-1&-1&0\\1&1&1&0\\1&0&1&0\end{array}\right]$
To test for linear independence, the easiest way is to evaluate $\displaystyle \det\left(\left[\begin{array}{ccc}0&-1&-1\\1&1&1\\1&0&1\end{array}\right]\right)$
This equals $\displaystyle \left|\begin{matrix}-1&-1\\1&1\end{matrix}\right|+\left|\begin{matrix}0&-1\\1&1\end{matrix}\right|=0+1=1\neq0$
Thus, they are linearly independent. Since this set spans $\displaystyle P_2$ and are linearly independent, they form a basis for $\displaystyle P_2$
Does this make sense?
--Chris