Solution LT.C41 Chris Black

First, we need to write the standard basis vectors $\vect{e}_1$ and $\vect{e}_2$ as linear combinations of $\colvector{2\\3}$ and $\colvector{3\\4}$. Starting with $\vect{e}_1$, we see that $\vect{e}_1 = -4\colvector{2\\3} + 3\colvector{3\\4}$, so we have \begin{align*} \lt{T}{\vect{e}_1} &= \lt{T}{-4\colvector{2\\3} + 3\colvector{3\\4}} = -4\,\lt{T}{\colvector{2\\3}} + 3 \,\lt{T}{\colvector{3\\4}}\\ &= -4\colvector{2\\2\\1} + 3\colvector{-1\\0\\2} = \colvector{-11\\-8\\2}. \end{align*} Repeating the process for $\vect{e}_2$, we have $\vect{e}_2 = 3\colvector{2\\3} - 2\colvector{3\\4}$, and we then see that \begin{align*} \lt{T}{\vect{e}_2} &= \lt{T}{3\colvector{2\\3} -2 \colvector{3\\4}} = 3\,\lt{T}{\colvector{2\\3}} - 2 \,\lt{T}{\colvector{3\\4}}\\ &= 3\colvector{2\\2\\1} -2 \colvector{-1\\0\\2} = \colvector{8\\6\\-1}. \end{align*} Thus, the matrix representation of $T$ is $A_T = \begin{bmatrix} -11 & 8\\ -8 & 6\\2 & -1 \end{bmatrix}$.