Q:

Matrix involving values of polynomials

I've been doing this problem but im stuck.
Be $f_1 f_2 f_3 \in \mathbb{R}_2$[$x  $]. Proove that {$f_1$,$ f_2$,$ f_3$} form a base of $\mathbb{R}_2$[$x  $] as $\mathbb{R}$ vector space, if and only if in $\mathbb{R}^{3x3}$ this matrix is invertible:
$ \begin{bmatrix} f_1(1) & f_1(2) & f_1(3)  \\ f_2(1) & f_2(2) & f_2(3) \\ f_3(1) & f_3(2) & f_3(3)\end{bmatrix}$ 
my attempt:  $\leftarrow$) i want to proove that $ \lambda_1 f_1 +\lambda_2  f_2 +\lambda_3 f_3 = 0 $ then $ \lambda_i   = 0 $   $  $    $ \forall i$ so they're linearly independant and then form a base. first i transpose the matrix, because i know that if a square matrix is invertible its transpose is too. Now i know that if i multiply the matrix by a vector ($ \lambda_1 \lambda_2 \lambda_3 $ ) and ask for the homogeneous solution i get that this ocurr if and only if $\lambda_i = 0 $ for $i= 1,2,3$ Then they're linearly independant and form a base.$ \\ $
$\\ \rightarrow$)...

A:

$\Rightarrow$ By contrapositive: suppose the matrix is not invertible — in this case, its transpose is not either, and there exists $\begin{pmatrix} a \\b \\c\end{pmatrix}\neq \begin{pmatrix} 0 \\0 \\0\end{pmatrix}$ such that
$$
\begin{pmatrix} f_1(1) & f_2(1) & f_3(1)  \\ f_1(2) & f_2(2) & f_3(2) \\ f_1(3) & f_2(3) & f_3(3)\end{pmatrix}\begin{pmatrix} a \\b \\c\end{pmatrix}=\begin{pmatrix} 0 \\0 \\0\end{pmatrix}
$$
i.e.
$$
\begin{align*}
a f_1(1) + b f_2(1) + c f_3(1) &= 0  \\
a f_1(2) + b f_2(2) + c f_3(2) &= 0  \\
a f_1(3) + b f_2(3) + c f_3(3) &= 0
\end{align*}
$$
En posant $P\stackrel{\rm{}def}{=}af_1+bf_2+cf_3\in\mathbb{R}_2[X]$, we have
$$
P(1)=P(2)=P(3)=0
$$ 
and therefore $P=0$. But since by assumption $(a,b,c)\neq(0,0,0)$, it means that $(f_1,f_2,f_3 )$ is linearly dependent, thus not a basis of $\mathbb{R}_2[X]$.

