1. **State the problem:** We are given a transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ defined by $T(c,b) = (c + b, -b, ab + 1)$ and asked to determine if $T$ is a linear transformation. 2. **Recall the definition of linear transformation:** A transformation $T$ is linear if for all vectors $\mathbf{u}, \mathbf{v}$ in the domain and all scalars $k$, the following hold: $$T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$$ $$T(k\mathbf{u}) = kT(\mathbf{u})$$ 3. **Check if $T$ satisfies these properties:** Let $\mathbf{u} = (c_1, b_1)$ and $\mathbf{v} = (c_2, b_2)$. Calculate $T(\mathbf{u} + \mathbf{v}) = T(c_1 + c_2, b_1 + b_2)$: $$T(c_1 + c_2, b_1 + b_2) = ((c_1 + c_2) + (b_1 + b_2), -(b_1 + b_2), a(c_1 + c_2)(b_1 + b_2) + 1)$$ Calculate $T(\mathbf{u}) + T(\mathbf{v})$: $$T(c_1, b_1) + T(c_2, b_2) = (c_1 + b_1, -b_1, a c_1 b_1 + 1) + (c_2 + b_2, -b_2, a c_2 b_2 + 1)$$ $$= (c_1 + b_1 + c_2 + b_2, -b_1 - b_2, a c_1 b_1 + a c_2 b_2 + 2)$$ 4. **Compare the third components:** From $T(\mathbf{u} + \mathbf{v})$ third component: $$a(c_1 + c_2)(b_1 + b_2) + 1 = a(c_1 b_1 + c_1 b_2 + c_2 b_1 + c_2 b_2) + 1$$ From $T(\mathbf{u}) + T(\mathbf{v})$ third component: $$a c_1 b_1 + a c_2 b_2 + 2$$ These are equal only if: $$a(c_1 b_2 + c_2 b_1) + 1 = 2$$ which is not true for all $c_1, c_2, b_1, b_2$. 5. **Check the zero vector condition:** For linearity, $T(0,0)$ must be $0$. Calculate: $$T(0,0) = (0 + 0, -0, a \cdot 0 \cdot 0 + 1) = (0, 0, 1) \neq (0,0,0)$$ 6. **Conclusion:** Since $T(0,0) \neq 0$ and the additivity property fails, $T$ is **not** a linear transformation. **Final answer:** $T$ is not linear because it does not satisfy the properties of linearity, specifically $T(0) \neq 0$ and additivity fails.