Let $t_1,t_2,t_3$ be distinct real numbers and consider the linear map
$T : \mathbb{R}[x]_{\leq 2} \rightarrow \mathbb{R}^3, \quad \quad p(x) \mapsto \begin{pmatrix}p(t_1)\\p(t_2)\\p(t_3)\end{pmatrix}$
I want to show that $T$ is an isomorphism. Since it is given that $T$ is linear, I just need to show that $T$ is bijective. My initial approach was to solve the system
$ \begin{pmatrix} a + bt_1 + ct_1^2\\ a + bt_2 + ct_2^2\\ a + bt_3 + ct_3^2 \end{pmatrix} = \begin{pmatrix} r_1\\ r_2\\ r_3 \end{pmatrix} $
for $a,b,c$ in terms of $r_1,r_2,r_3$, thus determining the inverse map $T^{-1}$ mapping a point in $\mathbb{R}^3$ to a polynomial of degree $\leq 2$.
I solved this system using Maple, and got a solution which was defined when $t_1,t_2,t_3$ where not all equal, which is fine by their definition, but I am wondering if there is a nicer argument, especially since the solution is rather ugly.
(this is not homework)