\[ if \ \ T: C^2 \longrightarrow \ C^2 \ satisfies \ \ T \Big ( \begin{bmatrix} 2 \\ 1 \end{bmatrix} \Big ) = \begin{bmatrix} 3 \\ 4 \end{bmatrix} \ and \ \ T \Big ( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \Big ) = \begin{bmatrix} -1 \\ 2 \end{bmatrix},\ find\ \ T\ \Big ( \begin{bmatrix} 4 \\ 3 \end{bmatrix} \Big ) \]
We start by assembling the column matrix A from \(u_i \ where \ T(u_i)=v_i\)
\[ A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \] We verify that \(B = \Big \{ \begin{bmatrix} 2 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 1 \end{bmatrix} \Big \}\) is a basis for \(C^2\) by showing that A is non-singular.
\[ A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 2 & 1 \\ 0 & 1/2 \end{bmatrix} \longrightarrow \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] The row-reduction shows us that the only solution to the homogenous equation is the trivial solution, so we conclude that A is non-singular and that the columns of A are a basis for \(C^2\)
So we now solve for values of \(v_1, v_2\) that make the system below true:
\[ \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 4\\ 3 \end{bmatrix} \] We augment A and row-reduce:
\[ \begin{bmatrix} 2 & 1 & 4 \\ 1 & 1 & 3 \end{bmatrix} \longrightarrow \begin{bmatrix} 2 & 1 & 4 \\ 0 & 1 & 2 \end{bmatrix} \longrightarrow \begin{bmatrix} 2 & 0 & 2 \\ 0 & 1 & 2 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} \] \[v_1 = 1, \ v_2 = 2\]
We can then write \(T \Big ( \begin{bmatrix} 4 \\ 3 \end{bmatrix} \Big )\) as:
\[ T \Big ( \begin{bmatrix} 4 \\ 3 \end{bmatrix} \Big ) = T \Big ( (1) \begin{bmatrix} 2 \\ 1 \end{bmatrix} + (2) \begin{bmatrix} 1 \\ 1 \end{bmatrix} \Big) = T \Big ( \begin{bmatrix} 2 \\ 1 \end{bmatrix} \Big ) + 2\ T \Big ( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \Big) \] and substitute \[ T \Big ( \begin{bmatrix} 4 \\ 3 \end{bmatrix} \Big ) = \begin{bmatrix} 3 \\ 4 \end{bmatrix} + 2\begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 8 \end{bmatrix} \] \[ T \Big ( \begin{bmatrix} 4 \\ 3 \end{bmatrix} \Big ) = \begin{bmatrix} 1 \\ 8 \end{bmatrix} \]