1. **Problem Statement:** Given a linear operator $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(3,1) = (2,-4)$ and $T(1,2) = (0,2)$, find $T(a,b)$ and $T(7,9)$. 2. **Recall:** Any vector $(a,b)$ in $\mathbb{R}^2$ can be expressed as a linear combination of the basis vectors $(3,1)$ and $(1,2)$: $$ (a,b) = x(3,1) + y(1,2) $$ where $x,y$ are scalars. 3. **Set up equations:** From the components, $$ 3x + y = a $$ $$ x + 2y = b $$ 4. **Solve for $x$ and $y$:** Multiply the second equation by 3: $$ 3x + 6y = 3b $$ Subtract the first equation: $$ (3x + 6y) - (3x + y) = 3b - a $$ $$ 5y = 3b - a $$ $$ y = \frac{3b - a}{5} $$ 5. Substitute $y$ back into the first equation: $$ 3x + \frac{3b - a}{5} = a $$ $$ 3x = a - \frac{3b - a}{5} = \frac{5a - (3b - a)}{5} = \frac{6a - 3b}{5} $$ $$ x = \frac{6a - 3b}{15} = \frac{2a - b}{5} $$ 6. **Find $T(a,b)$:** Using linearity, $$ T(a,b) = x T(3,1) + y T(1,2) = x(2,-4) + y(0,2) = (2x, -4x) + (0, 2y) = (2x, 2y - 4x) $$ Substitute $x$ and $y$: $$ T(a,b) = \left(2 \cdot \frac{2a - b}{5}, 2 \cdot \frac{3b - a}{5} - 4 \cdot \frac{2a - b}{5} \right) = \left( \frac{4a - 2b}{5}, \frac{6b - 2a - 8a + 4b}{5} \right) = \left( \frac{4a - 2b}{5}, \frac{10b - 10a}{5} \right) = \left( \frac{4a - 2b}{5}, 2b - 2a \right) $$ 7. **Calculate $T(7,9)$:** $$ T(7,9) = \left( \frac{4 \cdot 7 - 2 \cdot 9}{5}, 2 \cdot 9 - 2 \cdot 7 \right) = \left( \frac{28 - 18}{5}, 18 - 14 \right) = \left( 2, 4 \right) $$ **Final answers:** $$ T(a,b) = \left( \frac{4a - 2b}{5}, 2b - 2a \right), \quad T(7,9) = (2,4) $$