If \(T: C^2 \rightarrow C^3\) satisfies \(T(\begin{bmatrix} 2\\ 1\\ \end{bmatrix}) = \begin{bmatrix} 3 \\ 4 \end{bmatrix}\) & \(T(\begin{bmatrix} 1 \\ 1\\ \end{bmatrix}) = \begin{bmatrix} -1 \\ 2 \\ \end{bmatrix}\), find \(T(\begin{bmatrix} 4 \\ 3\\ \end{bmatrix})\).
Let’s express \(T(\begin{bmatrix} 4 \\ 3\\ \end{bmatrix})\) as a linear combination of \(\begin{bmatrix} 2 \\ 1 \\ \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}\).
We need to find coefficients such that \(\begin{bmatrix}4 \\ 3 \\ \end{bmatrix}\) can be expressed as a linear combination of \(\begin{bmatrix} 2 \\ 1 \\ \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}\).
\[ \begin{align*} a \begin{bmatrix} 2\\ 1\\ \end{bmatrix} + b \begin{bmatrix} 1\\ 1\\ \end{bmatrix} &= \begin{bmatrix} 4\\ 3\\ \end{bmatrix}\\ 2a + b &= 4 \quad \text{and} \quad a + b = 3\\ b &= 3-a\\ 2a + (3-a) &= 4 \quad \Rightarrow \quad a = 1 \\ \text{Substitute } a &= 1: \quad b = 3 - a \Rightarrow b = 3 - 1 = 2\\ T\left( \begin{bmatrix} 4\\ 3\\ \end{bmatrix} \right) &= 1 \cdot T\left( \begin{bmatrix} 2\\ 1\\ \end{bmatrix} \right) + 2 \cdot T\left( \begin{bmatrix} 1\\ 1\\ \end{bmatrix} \right)\\ \text{Substitute } T\left( \begin{bmatrix} 4\\ 3\\ \end{bmatrix} \right) &= 1 \cdot \begin{bmatrix} 3\\ 4\\ \end{bmatrix} + 2 \cdot \begin{bmatrix} -1\\ 2\\ \end{bmatrix} \\ &= \begin{bmatrix} 3\\ 4\\ \end{bmatrix} + \begin{bmatrix} -2\\ 4\\ \end{bmatrix} \\ &= \begin{bmatrix} 1\\ 8\\ \end{bmatrix} = T( \begin{bmatrix} 4\\ 3\\ \end{bmatrix}) \end{align*} \]
The solution is: \[ \begin{align*} T(\begin{bmatrix} 4\\ 3\\ \end{bmatrix}) = \begin{bmatrix} 1\\ 8\\ \end{bmatrix} \end{align*} \]