In \({{\rm{P}}_2}\), find the change-of-coordinates matrix from the basis \(B = \left\{ {1 - 3t,\,2 + t - 5{t^2},\,1 + 2t} \right\}\) to the standard basis. Then write \({t^2}\) as a linear combination of \(B\).

The \(C\)-coordinate vectors of \({b_1}\),\({b_2}\),\({b_3}\) are\({\left( {{b_1}} \right)_c} = \left( \begin{array}{l}\,\,\,\,\,\,1\,\\\,\,\,\,\,\,0\\\,\,\, - 3\end{array} \right)\), \({\left( {{b_2}} \right)_c} = \left( \begin{array}{l}\,\,\,2\\\,\,\,1\\ - 5\end{array} \right)\), \({\left( {{b_3}} \right)_c} = \left( \begin{array}{l}\,\,\,1\,\\\,\,\,2\\\,\,\,0\end{array} \right)\). Thus, the \(\mathop P\limits_{C \to B} \) matrix can be written as \(\mathop P\limits_{C \to B} = \left( {\begin{array}{*{20}{c}}1&2&1\\0&1&2\\3&{ - 5}&0\end{array}} \right)\).

If \(x\)is the vector \(\left\{ {{t^2}} \right\}\), then according to the result \({\left( x \right)_c} = \mathop P\limits_{C \to B} {\left( x \right)_B}\), you can write

\(\mathop P\limits_{C \to B} {\left( x \right)_B} = \left( \begin{array}{l}0\\0\\1\end{array} \right)\).

\(\left( {\begin{array}{*{20}{c}}1&2&1&0\\0&1&2&0\\{ - 3}&{ - 5}&0&1\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&3\\0&1&0&{ - 2}\\0&0&0&1\end{array}} \right)\)

\({\left( x \right)_B} = \left( \begin{array}{l}\,\,\,3\\ - 2\\\,\,\,1\end{array} \right)\)

\({t^2} = 3\left( {1 - 3{t^2}} \right) - 2\left( {2 + t - 5{t^2}} \right) + \left( {1 + 2t} \right)\)