Linear algebra homework problem involving basis and dual basis. Please help me get started on this problem:

Let $V = R^3$, and define $f_1, f_2, f_3 ∈ V^*$ as follows:
$f_1(x,y,z) = x - 2y$
$f_2(x,y,z) = x + y + z$
$f_3(x,y,z) = y-3z$
Prove that $\{f_1,f_2,f_3\}$ is a basis for $V^*$, and then find a basis for $V$ for which it is the dual basis.

 A: $\bf Hint:$ Suppose that $a f_1(x,y,z)+b f_2(x,y,z)+c f_3(x,y,z)=0$. Try different values for $(x,y,z)$, for example, if you substitute $(2,1,-3)$. We obtain $f_1(2,1,-3)=0$, $f_2(2,1,-3)=0$ and $f_3(2,1,-3)=10$, hence   $cf_3(2,1,-3)=10c$ which implies $c=0$.
A: Proving $\lbrace f_1,f_2,f_3\rbrace$ is a basis for $V^*$ can be done by row reducing the coefficients of $\lbrace f_1,f_2,f_3 \rbrace$ and showing that it has a rank of $3$.
The dual basis of $ \lbrace f_1,f_2,f_3 \rbrace $ is found by calculating the inverse of coefficients of $f_i$ which is:
$x_{1}= 
\
\begin{bmatrix}
  4/10 \\ -3/10 \\ -1/10 
\end{bmatrix},
x_{2}=
\begin{bmatrix}
  6/10 \\ 3/10 \\ 1/10 
\end{bmatrix},
x_{3}=
\begin{bmatrix}
  2/10 \\ 1/10 \\ -3/10 
\end{bmatrix}
]\
$
Checking that $f_{i}(x_{j})=$ $1,$ if $i=j$ and $0$ if $i \ne j $
$f_{1}(x_1)=(4/10)-2(-3/10)=1\\
f_{1}(x_2)=(6/10)-2(3/10)=0\\
f_{1}(x_3)=(2/10)-2(1/10)=0$
$f_{2}(x_1)=(4/10)+(-3/10)+(-1/10)=0\\
f_{2}(x_2)=(6/10)+(3/10)+(1/10)=1\\
f_{2}(x_3)=(2/10)+(1/10)+(-3/10)=0$
$f_{3}(x_1)=(-3/10)-3(-1/10)=0\\
f_{3}(x_2)=(3/10)-3(1/10)=0\\
f_{3}(x_3)=(1/10)-3(-3/10)=1$
