If \(T: \mathbb{C}^{2} -> \mathbb{C}^{2}\)
satisfies
\(T{({\lbrack\begin{array}{c}2\\[4pt]1\end{array}\rbrack})}={\lbrack\begin{array}{c}3\\[4pt]4\end{array}\rbrack} and T{({\lbrack\begin{array}{c}1\\[4pt]1\end{array}\rbrack})}={\lbrack\begin{array}{c}-1\\[4pt]2\end{array}\rbrack}\)
, find \(T{({\lbrack\begin{array}{c}4\\[4pt]3\end{array}\rbrack})}\)
\[{\lbrack\begin{array}{c}4\\[4pt]3\end{array}\rbrack}={\lbrack\begin{array}{c}2\\[4pt]1\end{array}\rbrack}+2{\lbrack\begin{array}{c}1\\[4pt]1\end{array}\rbrack}\]
\[T\left ( \begin{bmatrix} 4\\ 3 \end{bmatrix} \right ) = T\left ( \begin{bmatrix} 2\\1 \end{bmatrix} + 2 \begin{bmatrix} 1\\ 1 \end{bmatrix}\right )\]
\[T\left ( \begin{bmatrix} 2\\ 1 \end{bmatrix} \right ) + 2T\left ( \begin{bmatrix} 1\\1 \end{bmatrix} \right )\]
\[= \begin{bmatrix} 3\\ 4 \end{bmatrix} + 2 \begin{bmatrix} -1\\2 \end{bmatrix}\]
\[= \begin{bmatrix} 1\\ 8 \end{bmatrix}\]