Answer
The coordinate vector is;
$\begin{bmatrix}{c_1}\\{c_2}\\c_{3}\end{bmatrix} = \begin{bmatrix}2\\6\\-1\end{bmatrix}$
Work Step by Step
Basis for $P_{2}$
$B = \left\{ {1 + {t^2},t + {t^2},1 + 2t + {t^2}} \right\}$
We are required to find the coordinate vector of ${\bf{p}}(t) = 1 + 4t + 7{t^2}$ relative
to $B$
Let $[x]$ be the coordinate vector;
$[x]=\begin{bmatrix}{c_1}\\{c_2}\\c_{3}\end{bmatrix}$
$xB=P$
$\begin{array}{l} {c_1} \times \left( {1 + {t^2}} \right) + {c_2} \times \left( {t + {t^2}} \right) + {c_3} \times \left( {1 + 2t + {t^2}} \right) = 1 + 4t + 7{t^2}\\{c_1} \times \left( {1 + 0 \times t + {t^2}} \right) + {c_2} \times \left( {0 + t + {t^2}} \right) + {c_3} \times \left( {1 + 2t + {t^2}} \right) = 1 + 4t + 7{t^2}\\\left( {{c_1} + 0\cdot{c_2} + {c_3}} \right) + \left( {0\cdot{c_1} + {c_2} + 2{c_3}} \right)t + \left( {{c_1} + {c_2} + {c_3}} \right){t^2} = 1 + 4t + 7{t^2}\end{array}$
by taking the corresponding values on both sides of the equation we have;
$c_{1}+c_{3}=1$
$c_{2}+2c_{3}=4$
$c_{1}+c_{2}+c_{3}=7$
We then form the Matrix from the equations;
$=\begin{bmatrix}1&0&1\\0&1&2\\1&1&1\end{bmatrix}\begin{bmatrix}{c_1}\\{c_2}\\c_{3}\end{bmatrix}=\begin{bmatrix}1\\4\\7\end{bmatrix}$
Forming augmented matrix and row reducing;
$\begin{bmatrix}1&0&1&1\\0&1&2&4\\1&1&1&7\end{bmatrix}\sim\begin{bmatrix}1&0&0&2\\0&1&0&6\\0&0&1&-1\end{bmatrix}$
The coordinate vector is;
$[x]=\begin{bmatrix}{c_1}\\{c_2}\\c_{3}\end{bmatrix} = \begin{bmatrix}2\\6\\-1\end{bmatrix}$
