1. **Problem statement:** (a) Given a linear transformation $f : \mathbb{R}^2 \to \mathbb{R}^3$ with $$f\begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}3 \\ -1 \\ 3\end{pmatrix}, \quad f\begin{pmatrix}0 \\ 1\end{pmatrix} = \begin{pmatrix}-4 \\ 2 \\ -5\end{pmatrix},$$ find $f\begin{pmatrix}-7 \\ 3\end{pmatrix}$. (b) Given a linear transformation $f : \mathbb{R}^{12} \to \mathbb{R}^2$ with $$f(\vec{e}_4) = \begin{pmatrix}2 \\ -2\end{pmatrix}, \quad f(\vec{e}_7) = \begin{pmatrix}4 \\ -5\end{pmatrix}, \quad f(\vec{e}_8) = \begin{pmatrix}4 \\ -3\end{pmatrix},$$ find $$f(5\vec{e}_4 + 6\vec{e}_7) - f(2\vec{e}_8 + 4\vec{e}_7).$$ (c) Given a linear transformation $T : V \to \mathbb{R}^2$ with $$T(\vec{v}_1) = \begin{pmatrix}-2 \\ -5\end{pmatrix}, \quad T(\vec{v}_2) = \begin{pmatrix}-1 \\ 3\end{pmatrix}, \quad T(\vec{v}_3) = \begin{pmatrix}-4 \\ 5\end{pmatrix},$$ find $$-3T(\vec{v}_1) + T(3\vec{v}_2 + 6\vec{v}_3).$$ 2. **Formula and rules:** For linear transformations $f$ and $T$, the key property is linearity: $$f(a\vec{x} + b\vec{y}) = a f(\vec{x}) + b f(\vec{y}),$$ $$T(a\vec{u} + b\vec{w}) = a T(\vec{u}) + b T(\vec{w}),$$ where $a,b$ are scalars and $\vec{x}, \vec{y}, \vec{u}, \vec{w}$ are vectors. 3. **Step-by-step solutions:** **(a)** $$f\begin{pmatrix}-7 \\ 3\end{pmatrix} = -7 f\begin{pmatrix}1 \\ 0\end{pmatrix} + 3 f\begin{pmatrix}0 \\ 1\end{pmatrix} = -7 \begin{pmatrix}3 \\ -1 \\ 3\end{pmatrix} + 3 \begin{pmatrix}-4 \\ 2 \\ -5\end{pmatrix}.$$ Calculate each component: $$= \begin{pmatrix}-21 \\ 7 \\ -21\end{pmatrix} + \begin{pmatrix}-12 \\ 6 \\ -15\end{pmatrix} = \begin{pmatrix}-33 \\ 13 \\ -36\end{pmatrix}.$$ **(b)** Use linearity: $$f(5\vec{e}_4 + 6\vec{e}_7) - f(2\vec{e}_8 + 4\vec{e}_7) = 5 f(\vec{e}_4) + 6 f(\vec{e}_7) - 2 f(\vec{e}_8) - 4 f(\vec{e}_7).$$ Group terms: $$= 5 \begin{pmatrix}2 \\ -2\end{pmatrix} + (6 - 4) \begin{pmatrix}4 \\ -5\end{pmatrix} - 2 \begin{pmatrix}4 \\ -3\end{pmatrix} = \begin{pmatrix}10 \\ -10\end{pmatrix} + 2 \begin{pmatrix}4 \\ -5\end{pmatrix} - \begin{pmatrix}8 \\ -6\end{pmatrix}.$$ Calculate: $$= \begin{pmatrix}10 \\ -10\end{pmatrix} + \begin{pmatrix}8 \\ -10\end{pmatrix} - \begin{pmatrix}8 \\ -6\end{pmatrix} = \begin{pmatrix}10 + 8 - 8 \\ -10 - 10 + 6\end{pmatrix} = \begin{pmatrix}10 \\ -14\end{pmatrix}.$$ **(c)** Apply linearity: $$-3 T(\vec{v}_1) + T(3 \vec{v}_2 + 6 \vec{v}_3) = -3 \begin{pmatrix}-2 \\ -5\end{pmatrix} + 3 T(\vec{v}_2) + 6 T(\vec{v}_3).$$ Calculate each term: $$= \begin{pmatrix}6 \\ 15\end{pmatrix} + 3 \begin{pmatrix}-1 \\ 3\end{pmatrix} + 6 \begin{pmatrix}-4 \\ 5\end{pmatrix} = \begin{pmatrix}6 \\ 15\end{pmatrix} + \begin{pmatrix}-3 \\ 9\end{pmatrix} + \begin{pmatrix}-24 \\ 30\end{pmatrix}.$$ Sum components: $$= \begin{pmatrix}6 - 3 - 24 \\ 15 + 9 + 30\end{pmatrix} = \begin{pmatrix}-21 \\ 54\end{pmatrix}.$$ 4. **Final answers:** (a) $f\begin{pmatrix}-7 \\ 3\end{pmatrix} = \begin{pmatrix}-33 \\ 13 \\ -36\end{pmatrix}$ (b) $f(5\vec{e}_4 + 6\vec{e}_7) - f(2\vec{e}_8 + 4\vec{e}_7) = \begin{pmatrix}10 \\ -14\end{pmatrix}$ (c) $-3 T(\vec{v}_1) + T(3 \vec{v}_2 + 6 \vec{v}_3) = \begin{pmatrix}-21 \\ 54\end{pmatrix}$