Find a basis $\alpha = \{ \vec\alpha_1, \vec\alpha_2, \vec\alpha_3 \}$ of $P_2(R)$, such that $[2 + 5x + 4x^2]_\alpha =\left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right)$
I have approached the problem the following way: $\begin{align*} \vec v &= 2 + 5x + 4x^2\\ s &= \{1, x, x^2\}\\ [\vec v]_s &= \left(\begin{matrix} 2 \\ 5 \\ 4 \end{matrix}\right)\\ [\vec v]_\alpha &= \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right)\\ _sC_\alpha &=\left( \begin{matrix} \uparrow & \uparrow & \uparrow \\ \alpha_1 & \alpha_2 & \alpha_3 \\ \downarrow & \downarrow & \downarrow \end{matrix} \right) \end{align*}$
Therefore: $\begin{align*} {_sC_\alpha}^{-1} [\vec v]_s &= [\vec v]_\alpha\\ [\vec v]_s &= _sC_\alpha [\vec v]_\alpha \end{align*}$ So $ \left(\begin{matrix} 2 \\ 5 \\ 4 \end{matrix}\right) = \left(\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right) \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right)$ For which one possible solution is $\left(\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right) = \left(\begin{matrix} 2 & 0 & 0 \\ 0 & 5/2 & 0 \\ 0 & 0 & 4/3 \end{matrix}\right)$
My question basically boils down to this: Is this a valid answer? It appears to me like under the Field R there exist infinitely many such bases.
PS: Is it just me or does the latex formatting for \bordermatrix not work on this site?