Question 4.1: Let T:R² → R³ be the linear transformation defined by T([x1 ......
Let T:R² → R³ be the linear transformation defined by T \left\lgroup\left[\begin{array}{l}{x_{1}}\\ {x_{2}}\end{array}\right] \right\rgroup =\left[\begin{array}{c}{{x_{2}}}\\ {{-5x_{1}+13x_{2}}}\\ {{-7x_{1}+16x_{2}}}\end{array}\right].
Find the matrix of the transformation T with respect to the bases S_{1}=\{v_{1},v_{2}\} for R² and S_{2}=\{w_{1},w_{2},w_{3}\} for R³, where v_{1}={\left[\begin{array}{c}{3}\\ {1}\end{array}\right]},\ v_{1}={\left[\begin{array}{c}{5}\\ {2}\end{array}\right]},\,w_{1}={\left[\begin{array}{c}{1}\\ {0}\\ {-1}\end{array}\right]},w_{2}={\left[\begin{array}{c}{-1}\\ {2}\\ {2}\end{array}\right]},w_{3}={\left[\begin{array}{c}{0}\\ {1}\\ {2}\end{array}\right]}.
Here, T(\nu_{1})=T\left\lgroup\left[\begin{array}{c}{{3}}\\ {{1}}\end{array}\right]\right\rgroup=\left[\begin{array}{c}{{1}}\\ {{-5(3)+13(1)}}\\ {{-7(3)+16(1)}}\end{array}\right]=\left[\begin{array}{c}{{1}}\\ {{-2}}\\-5\end{array}\right]
\\ \qquad\qquad T(\nu_{2})=T{\left\lgroup\left[\begin{array}{c}{5}\\ {2}\end{array}\right]\right\rgroup}={\left[\begin{array}{c}{2}\\ {-5(5)+13(2)}\\ {-7(5)+16(2)}\end{array}\right]}={\left[\begin{array}{c}{2}\\ {1}\\ {-3}\end{array}\right]}Let us express \nu_1 as a linear combination of w_1,w_2,w_3
T(\nu_{1})=k_{1}w_{1}+k_{2}w_{2}+k_{3}w_{3}.\therefore\left[\begin{array}{c}{{1}}\\-2\\ {{-5}}\end{array}\right]=k_{1}\left[\begin{array}{c}{{1}}\\{{0}}\\ {{-1}}\end{array}\right]+k_{2}\left[\begin{array}{c}{{-1}}\\ {{2}}\\2\end{array}\right]+k_{3}\left[\begin{array}{c}{{0}}\\ {{1}}\\ {{2}}\end{array}\right].
\therefore\left[\begin{array}{c}{{1}}\\-2\\ {{-5}}\end{array}\right]=\left[\begin{array}{c}{{k_{1}-k_{2}}}\\ {{2k_{2}+k_{3}}}\\ {{-k_{1}+2k_{2}+2k_{3}}}\end{array}\right].
Comparing both sides,
k_{1}-k_{2}=1,2k_{2}+k_{3}=-2,-k_{1}+2k_{2}+2k_{3}=-5.Using Gauss elimination method,
\begin{array}{c}{k_{1}=1,k_{2}=0,k_{3}=-2}\\[0.5 em] {T(v_{1})=w_{1}-2w_{3}}\\[0.5 em] {\pmb{\big[}T(v_{1})\pmb{\big]}_{s_2}=\left[\begin{array}{c}{{1}}\\ {{0}}\\ {{-2}}\end{array}\right].}\end{array}Express \nu_2 as a linear combination of w_1,w_2,w_3
T(\nu_{2})=k_{1}w_{1}+k_{2}w_{2}+k_{3}w_{3}.\therefore \left[{\begin{array}{c}{2}\\ {1}\\ {-3}\end{array}}\right]=k_{1}\left[{\begin{array}{c}{1}\\ {0}\\ {-1}\end{array}}\right]+k_{2}\left[\begin{array}{c}{{-1}}\\ {{2}}\\ {{2}}\end{array}\right]+k_{3}\left[\begin{array}{c}{{0}}\\ {{1}}\\ {{2}}\end{array}\right].
\therefore\left[\begin{array}{c}{{2}}\\ {{1}}\\ {{-3}}\end{array}\right]=\left[\begin{array}{c}{{k_{1}-k_{2}}}\\{{2k_{2}+k_{3}}}\\ {{-k_{1}+2k_{2}+2k_{3}}}\end{array}\right].
k_{1}-k_{2}=2,2k_{2}+k_{3}=1,-k_{1}+2k_{2}+2k_{3}=-3.Using Gauss elimination method,
\begin{array}{r}{k_{1}=3,k_{2}=1,k_{3}=-1}\\[0.5 em] {T\bigl(\nu_{2}\bigr)=3w_{1}+w_{2}-w_{3}}\\[0.5 em] {\pmb{\big[}T(v_{2})\pmb{\big]} _{s_2}=\left[\begin{array}{c}3\\{1}\\ {-1}\end{array}\right].}\end{array}The matrix of the transformation T with respect to the bases S_1 and S_2 is
\pmb{\big[}T\pmb{\big]} _{s_{2},s_{1}}=\left[\pmb{\big[}T(v_{1})\pmb{\big]}_{s_{2}}| \pmb{\big[}T(v_{2})\pmb{\big]}_{s_{2}}\right]=\left[{\begin{array}{r r}{1}&{3}\\ {0}&{1}\\ {-2}&{-1}\end{array}}\right].