Answer
See answer below
Work Step by Step
Let $V$ be a vector space of problem 14.
Assume the set $S=\{(1,4);(2,1)\}$
Since $S \in V \rightarrow S \subset V$
By problem 14 we have $(0,1)$ is the zero vector of $V$.
Let $a$ and $b$ be scalars such that
$a(1,4)+b(2,1) \\
=(a,4^a)+(2b,2^b) \\
=(a+2b,4^a.2^b)\\
=(a+2b,(2^2)^a.2^b)\\
=(a+2b,2^{2a}.2^b)\\
=(a+2b,2^{2a+b})\\
=(0,1)$
Hence: $a+2b=0\\
2^{2a+b}=1 \\
\rightarrow a+2b=0\\
2a+b=0$
Thus $2(a+2b)-(2a+b)=0 \rightarrow 3b=0 \rightarrow b=0 \\
a+2(0)=0 \rightarrow a=0$
Hence, $S$ is a basic for the vector space $V$ given in problem 13.
