Answer
See below
Work Step by Step
Given: $V=R^4\\
S=\{(6,-3,2,0),(1,1,1,0),(1,-8,-1,0)\}$
We have $dim(R^4)=4$ but we can see $S$ just have 3 vectors.
Thus $S$ do not span $R^4$
Obtain:
$a(6,-3,2,0)+b(1,1,1,0)+c(1,-8,-1,0)=0$
$\rightarrow 6a+b+c=0\\
-3a+b-8c=0\\
2a+b-c=0$
From $6a+b+c=0 \rightarrow a=-\frac{a+b}{6}$
Substitute: $-3(-\frac{a+b}{6})+c-8c=0\\
2(-\frac{a+b}{6})+b-c=0\\
\rightarrow \frac{3b-15c}{2}=0\\
\frac{2b-4c}{3}=0\\
\rightarrow b=5c\\
\frac{2b-4c}{3}=0$
Substitute: $\frac{2.5c-4c}{3}=0 \rightarrow 2c =0 \rightarrow c=0$
then $a=0\\
b=0$
We have $rank\begin{bmatrix}
6 & -3 & 2 & 0\\
1 & 1 & 1 &0\\
1 & -8 & -1 & 0
\end{bmatrix}=3\\
\rightarrow rank(A)=dim(S)=3$
Consequently, the given set of vectors $S$ are linearly independent.
