Answer
See below
Work Step by Step
Let $S$ be the subspace of $R^2$ spanned by the vectors $v_1=(2,1)\\v_2=(3,1) \in R^2$.
a) Let $v=(x,y)\in R^2$
We can see that $(3y-x)(2,1)+(x-2y)(3,1)\\=(2(3y-x),3y-x)+(3(x-2y),x-2y)\\=(6y-2x+3x-6y,3y-x+x-2y)\\=(x,y)\\=v\\ \rightarrow v \in span\{v_1,v_2\}$
Hence, any vector in $R^2$ can be written as a linear combination of $v_1$ and $v_2$
Consequently, $\{v_1,v_2\}$ spans $R^2$
b) Let $a,b$ be scalars such that
$a(2,1)+b(3,1)=(0,0)$
Obtain $(2a+3b,a+b)=(0,0)\\\rightarrow 2a+3b=0,a+b=0$
then $a+b=0 \rightarrow a=-b\\
2(-b)+3b=0 \rightarrow b=0\\
\rightarrow a=0$
The linear combination of $v_1$ and $v_2$ is trivial and therefore set $\{v_1,v_2\}$ is linearly independent in $R^2$
c) From exercise b), $\{v_1,v_2\}$ is a basic for $R^2 \rightarrow \dim[R^2]=2 $ (by Corollary 4.6.13)
