Find \(T\left( {{{\bf{b}}_1}} \right)\), \(T\left( {{{\bf{b}}_2}} \right)\) and \(T\left( {{{\bf{b}}_3}} \right)\) for \(B = \left\{ {1,t,{t^2}} \right\}\) by using \(T\left( {{a_0} + {a_1}t + {a_2}{t^2}} \right) = 3{a_0} + \left( {5{a_0} - 2{a_1}} \right)t + \left( {4{a_1} + {a_2}} \right){t^2}\).
\(\begin{aligned}{c}T\left( {{{\bf{b}}_1}} \right) &= T\left( 1 \right)\\ &= 3 + 5t\end{aligned}\)
\(\begin{aligned}{c}T\left( {{{\bf{b}}_2}} \right) &= T\left( t \right)\\ &= - 2t + 4{t^2}\end{aligned}\)
\(\begin{aligned}{c}T\left( {{{\bf{b}}_3}} \right) &= T\left( {{t^2}} \right)\\ &= {t^2}\end{aligned}\)
Find \({\left( {T\left( {{{\bf{b}}_1}} \right)} \right)_B}\), \({\left( {T\left( {{{\bf{b}}_2}} \right)} \right)_B}\) and \({\left( {T\left( {{{\bf{b}}_3}} \right)} \right)_B}\).
\({\left( {T\left( {{{\bf{b}}_1}} \right)} \right)_B} = \left( {\begin{aligned}3\\5\\0\end{aligned}} \right)\), \({\left( {T\left( {{{\bf{b}}_2}} \right)} \right)_B} = \left( {\begin{aligned}0\\{ - 2}\\4\end{aligned}} \right)\), \({\left( {T\left( {{{\bf{b}}_3}} \right)} \right)_B} = \left( {\begin{aligned}0\\0\\1\end{aligned}} \right)\)
Form a matrix \(T\) for the obtained vectors by using the formula \({\left( {T\left( {\bf{x}} \right)} \right)_C} = \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_C}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_C}}& \cdots &{{{\left( {T\left( {{{\bf{b}}_n}} \right)} \right)}_C}}\end{aligned}} \right)\), where \(n = 3\).
\(\begin{aligned}{\left( {T\left( {\bf{x}} \right)} \right)_B} &= \left( {\begin{aligned}{*{20}{c}}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_B}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_B}}&{{{\left( {T\left( {{{\bf{b}}_3}} \right)} \right)}_B}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}3&{}&0&{}&0\\5&{}&{ - 2}&{}&0\\0&{}&4&{}&1\end{aligned}} \right)\end{aligned}\)
So, the required matrix is \(\left( {\begin{aligned}3&{}&0&{}&0\\5&{}&{ - 2}&{}&0\\0&{}&4&{}&1\end{aligned}} \right)\).
