Answer
The coordinate vector is;
$[x]=\begin{bmatrix}{c_1}\\{c_2}\\c_{3}\end{bmatrix} = \begin{bmatrix}3\\2\\-1\end{bmatrix}$
Work Step by Step
Basis for $P_{2}$
$B = \left\{ {1 - {t^2},t - {t^2},2 - 2t + {t^2}} \right\}$
We are required to find the coordinate vector of ${\bf{p}}(t) = 3 + t - 6{t^2}$ relative
to $B$
Let $[x]$ be the coordinate vector;
$[x]=\begin{bmatrix}{c_1}\\{c_2}\\c_{3}\end{bmatrix}$
$xB=P$
$\begin{array}{l} {c_1} \times \left( {1 - {t^2}} \right) + {c_2} \times \left( {t - {t^2}} \right) + {c_3} \times \left( { 2- 2t + {t^2}} \right) = 3 + t - 6{t^2}\}\end{array}$
by taking the corresponding values on both sides of the equation we have;
$c_{1}+2c_{3}=1$
$c_{2}-c_{3}=3$
$-c_{1}-c_{2}+c_{3}=-6$
We then form the Matrix from the equations;
$=\begin{bmatrix}1&0&2\\0&1&-1\\-1&-1&1\end{bmatrix}\begin{bmatrix}{c_1}\\{c_2}\\c_{3}\end{bmatrix}=\begin{bmatrix}1\\3\\-6\end{bmatrix}$
Forming augmented matrix and row reducing;
$\begin{bmatrix}1&0&2&1\\0&1&-1&3\\-1&-1&1&-6\end{bmatrix}\sim\begin{bmatrix}1&0&0&3\\0&1&0&2\\0&0&1&-1\end{bmatrix}$
The coordinate vector is;
$[x]=\begin{bmatrix}{c_1}\\{c_2}\\c_{3}\end{bmatrix} = \begin{bmatrix}3\\2\\-1\end{bmatrix}$
