Answer
See answer below
Work Step by Step
Assume that:
$p_0(x)=1$
$p_1(x)=x$
$p_2(x)=\frac{1}{2}(3x^2-1)$
We obtain:
$W(p_0,p_1,p_2)=\begin{bmatrix}
1 & x & \frac{1}{2}(3x^2-1)\\
0 & 1 & \frac{1}{2}6x \\
0 & 0 & 3
\end{bmatrix} =1 \times 1 \times 3=3$
$\rightarrow W(p_0,p_1,p_2)=3 \ne 0 $
The set of vectors $\{p_0,p_1,p_2\}$ is linearly independent.
Hence $\{p_1,p_2,p_3\}$ is a basis for $P_2$
