Let \(T\) be a linear transformation such that \[ T\bigg( \begin{bmatrix}2\\1\end{bmatrix}\bigg)= \begin{bmatrix}3\\4\end{bmatrix}\\ \text{and}\\ T\bigg(\begin{bmatrix}1\\1\end{bmatrix}\bigg)= \begin{bmatrix}-1\\2\end{bmatrix}\\ \] Find \[ T\bigg(\begin{bmatrix}4\\3\end{bmatrix}\bigg) \]
We know that a linear transformation of the linear combination is the same combination applied to the linear transformation. Since \[\begin{bmatrix}4\\3\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}+2\begin{bmatrix}1\\1\end{bmatrix} \] it follows that \[ T\bigg(\begin{bmatrix}4\\3\end{bmatrix}\bigg)=T\bigg(\begin{bmatrix}2\\1\end{bmatrix}+2\begin{bmatrix}1\\1\end{bmatrix}\bigg)\\ \text{so}\\ T\bigg(\begin{bmatrix}4\\3\end{bmatrix}\bigg)=T\bigg(\begin{bmatrix}2\\1\end{bmatrix}\bigg)+T\bigg(2\begin{bmatrix}1\\1\end{bmatrix}\bigg)\\ T\bigg(\begin{bmatrix}4\\3\end{bmatrix}\bigg)=T\bigg(\begin{bmatrix}2\\1\end{bmatrix}\bigg)+2T\bigg(\begin{bmatrix}1\\1\end{bmatrix}\bigg)\\ T\bigg(\begin{bmatrix}4\\3\end{bmatrix}\bigg)=\begin{bmatrix}3\\4\end{bmatrix}+2\begin{bmatrix}-1\\2\end{bmatrix}\\ T\bigg(\begin{bmatrix}4\\3\end{bmatrix}\bigg)=\begin{bmatrix}3\\4\end{bmatrix}+\begin{bmatrix}-2\\4\end{bmatrix}\\ T\bigg(\begin{bmatrix}4\\3\end{bmatrix}\bigg)=\begin{bmatrix}1\\8\end{bmatrix} \]