Answer
See answer below
Work Step by Step
We are given:
$S=\{5x^2,1+6x,-3-x^2\}$
We obtain:
$C_1(5x^2)+C_2(1+6x)+C_3(-3-x^2)=0$
$(C_2-3C_3)+x(6C_2)+x^2(5C_1-C_3)=0$
We have:
$C_2-3C_3=0$
$6C_2=0$
$5C_1-C_3=0$
$A=\begin{bmatrix}
5 & 0 & -1 | 0\\
0 & 6 & 0| 0 \\
0 & 1 & -3 | 0
\end{bmatrix}$
$\det A=-90\ne 0$
Since $\det A \ne 0 $, set $S$ of vector is linearly dependent and therefore, set $S$ is a basis of $P_2(R)$
