Finding Dual Basis from a basis in $\mathbb R^2$ I am given the fact that these two vectors form the basis B of $\mathbb R^2$:
$$
B=\{\begin{bmatrix} 2 \\ 1 \\ \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix}\}
$$
and then asked to find the dual basis or the basis of $(\mathbb R^2)^*$. 
I would really appreciate if anyone could explain the procedure to follow for these types of questions.

$(\mathbb{R}^2)^*$ consists of linear maps $\ell:\mathbb{R}^2\to \mathbb{R}$ which  have  standard matrix representation 
$$ \begin{bmatrix}
a&b
\end{bmatrix}$$
where $a=\ell(e_1)$ and $b=\ell(e_2)$ for $e_1,e_2$ the  standard basis. Let's write your basis vectors as $b_1,b_2$ respectively. The dual basis is the basis $(\phi_1,\phi_2)$ for $(\mathbb{R}^2)^*$ satisfying $\phi_i(b_j)=\delta_{ij}$ for $\delta_{ij}=1$ with $i=j$ and $0$ else. So, $\phi_1(b_1)=1$ and $\phi_1(b_2)=0$. Notice that $b_2-b_1=e_1$. So, 
$$\phi_1(e_1)=\phi_1(b_2-b_1)=\phi_1(b_2)-\phi_1(b_1)=-1.$$
Similarly, $3b_1-2b_2=e_2$ so
$$ \phi_1(e_2)=\phi_1(3b_1-2b_2)=3\phi_1(b_1)-2\phi_1(b_2)=3.$$
Hence, $\phi_1$ has matrix representation
$$ \begin{bmatrix}
-1&3
\end{bmatrix}.$$
Applying the same reasoning, we get 
$$ \phi_2(e_1)=\phi_2(b_2-b_1)=\phi_2(b_2)-\phi_2(b_1)=1$$
$$ \phi_2(e_2)=\phi_2(3b_1-2b_2)=3\phi_2(b_1)-2\phi_2(b_2)=-2.$$
Hence, $\phi_2$ has matrix representation 
$$ \begin{bmatrix}
1&-2
\end{bmatrix}.$$
A simple calculation reveals that indeed these $\phi_1,\phi_2$ satisfy $\phi_i(b_j)=\delta_{ij}$.
