1. **State the problem:** We need to show that the transformation $T : \mathbb{R}^3 \to \mathbb{R}^2$ defined by $T(x, y, z) = (z, x + y)$ is linear. 2. **Recall the definition of linearity:** A transformation $T$ is linear if for all vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^3$ and scalars $c \in \mathbb{R}$, the following hold: - $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ - $T(c\mathbf{u}) = cT(\mathbf{u})$ 3. **Check additivity:** Let $\mathbf{u} = (x_1, y_1, z_1)$ and $\mathbf{v} = (x_2, y_2, z_2)$. $$ T(\mathbf{u} + \mathbf{v}) = T(x_1 + x_2, y_1 + y_2, z_1 + z_2) = (z_1 + z_2, (x_1 + x_2) + (y_1 + y_2)) $$ $$ T(\mathbf{u}) + T(\mathbf{v}) = (z_1, x_1 + y_1) + (z_2, x_2 + y_2) = (z_1 + z_2, (x_1 + y_1) + (x_2 + y_2)) $$ Since $(x_1 + x_2) + (y_1 + y_2) = (x_1 + y_1) + (x_2 + y_2)$, additivity holds. 4. **Check homogeneity:** For scalar $c$ and vector $\mathbf{u} = (x, y, z)$, $$ T(c\mathbf{u}) = T(cx, cy, cz) = (cz, cx + cy) = c(z, x + y) = cT(\mathbf{u}) $$ 5. **Conclusion:** Both additivity and homogeneity hold, so $T$ is a linear transformation.