The coordinate vector of \displaystyle{\left[{\mathbf{{{p}}}}

Any arbitrary vector in ${\displaystyle {\mathbb{P}}_{\mathbb{2}}}$ can be written as,
${\displaystyle [\mathbf{p}\left(t\right)=a(1+t^2)+b(t+t^2)+c(1+2t+t^2)]}$
Thus, for the vector ${\displaystyle [\mathbf{p}\left(t\right)=6+3t-t^2]}$ can be written as,
1) ${\displaystyle [a(1+t)+b(1+t^2)+c(t+t^2)=6+3t-t^2]}$
On comparing the terms of both the sides in the above equation (1) gives the following equations.
${\displaystyle a+b=6}$
${\displaystyle a+c=3}$
${\displaystyle b+c=-1}$
The representation of the linear system in a matrix is,
$\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& 1\end{array}\right]\left[\begin{array}{c}a\\ c\\ d\end{array}\right]=\left[\begin{array}{c}6\\ 3\\ -1\end{array}\right]$
The solution of the above system gives as by solving the augmented matrix
$\left[\begin{array}{cccc}1& 1& 0& 6\\ 1& 0& 1& 3\\ 0& 1& 1& -1\end{array}\right]$
Use Elementary row transformations to reduce the augmented matrix to row reduced Echelon form.
Step 1:
$\left[\begin{array}{cccc}1& 1& 0& 6\\ 1& 0& 1& 3\\ 0& 1& 1& -1\end{array}\right]\underrightarrow{R_2\to R_2-R_1}\left[\begin{array}{cccc}1& 1& 0& 6\\ 0& -1& 1& -3\\ 0& 1& 1& -1\end{array}\right]$
Step 2
$\left[\begin{array}{cccc}1& 1& 0& 6\\ 0& -1& 1& -3\\ 0& 1& 1& -1\end{array}\right]\underrightarrow{R_3\to R_3+R_2}\left[\begin{array}{cccc}1& 1& 0& 6\\ 0& -1& 1& -3\\ 0& 0& 2& -4\end{array}\right]$
Step 3:
$\left[\begin{array}{cccc}1& 1& 0& 6\\ 0& -1& 1& -3\\ 0& 0& 2& -4\end{array}\right]\underrightarrow{R_1\to R_1+R_2-\frac{1}{2}{R}_3}\left[\begin{array}{cccc}1& 0& 0& 5\\ 0& -1& 1& -3\\ 0& 0& 2& -4\end{array}\right]$
Step 4:
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