Let T_{1}:R^{2}\to R^{2}\;{\mathrm{and}}\;T_{2}:R^{2}\to R^{3} be the transformations given by T_{1}\left(x,y\right)=\left(x+y,y\right) and T_{2}\left(x,y\right)=\left(2x,\,y,\,\,x+y\right). Find the formula for T_{2}\circ T_{1}\left(x,y\right).
Let the standard matrix for T_{1}\left(x,y\right)=\left(x+y,y\right) transformation is T_{1}=\left[\begin{array}{c c}{{1}}&{{1}}\\ {{0}}&{{1}}\end{array}\right] and T_{2}(x,y)=(2x,y,x+y) is T_{2}=\left[\begin{array}{c c c}{{2}}&{{0}}\\ {{0}}&{{1}}\\ {{1}}&{{1}}\end{array}\right].
We know that from the above composition formula T_{2}\circ T_{1}=T_{2}\cdot T_{1}
\begin{array}{r l}{={\left[\begin{array}{l l}{\,~2}&{\,~0}\\ {\,~0}&{\ \ 1}\\ {\ \ 1}&{\ \ 1\ }\end{array}\right]}}\end{array}\left[\begin{array}{l l}{1}&{1}\\ {0}&{1}\end{array}\right]
\\ \mathbf=\left[{\begin{array}{r r}{\ \ 2}&{2\ \ }\\ {0}&{1\ \ }\\ {1}&{2\ \ }\end{array}}\right]Therefore, T_{2}\circ T_{1}\left(x,y\right)=\left[{\begin{array}{r r}{2}&{\,\,2\ \ }\\ {\,0}&{\,1\ \ }\\ {\,\,\,\,1}&{\,\,2\ \ }\end{array}}\right]\left[\begin{array}{c}x\\ \ \ y\ \ \end{array}\right]
T_{2}\circ T_{1}\left(x,y\right)=\left(2x+2y,\ y,\ x+2y\right).