1. **State the problem:** We have a vector $\mathbf{x} = \begin{bmatrix} \frac{11}{7} \\ \frac{18}{11} \end{bmatrix}$ in $\mathbb{R}^2$ and two linear transformations: a rotation $R_\theta$ with $\theta = \frac{3\pi}{4}$ and a shear $Q_m$ with $m = \frac{13}{8}$. We want to find $R_\theta \mathbf{x}$ and $Q_m \mathbf{x}$ and graph them. 2. **Formulas:** - Rotation matrix: $$R_\theta = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$ - Shear matrix (horizontal shear): $$Q_m = \begin{bmatrix} 1 & m \\ 0 & 1 \end{bmatrix}$$ 3. **Calculate $R_\theta \mathbf{x}$:** - Compute $\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$ and $\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$. - So, $$R_\theta = \begin{bmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix}$$ - Multiply: $$R_\theta \mathbf{x} = \begin{bmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} \frac{11}{7} \\ \frac{18}{11} \end{bmatrix} = \begin{bmatrix} -\frac{\sqrt{2}}{2} \cdot \frac{11}{7} - \frac{\sqrt{2}}{2} \cdot \frac{18}{11} \\ \frac{\sqrt{2}}{2} \cdot \frac{11}{7} - \frac{\sqrt{2}}{2} \cdot \frac{18}{11} \end{bmatrix}$$ - Simplify each component: $$x_1 = -\frac{\sqrt{2}}{2} \left( \frac{11}{7} + \frac{18}{11} \right) = -\frac{\sqrt{2}}{2} \cdot \frac{11 \cdot 11 + 18 \cdot 7}{7 \cdot 11} = -\frac{\sqrt{2}}{2} \cdot \frac{121 + 126}{77} = -\frac{\sqrt{2}}{2} \cdot \frac{247}{77}$$ $$x_2 = \frac{\sqrt{2}}{2} \left( \frac{11}{7} - \frac{18}{11} \right) = \frac{\sqrt{2}}{2} \cdot \frac{11 \cdot 11 - 18 \cdot 7}{7 \cdot 11} = \frac{\sqrt{2}}{2} \cdot \frac{121 - 126}{77} = \frac{\sqrt{2}}{2} \cdot \left(-\frac{5}{77}\right) = -\frac{5 \sqrt{2}}{154}$$ 4. **Calculate $Q_m \mathbf{x}$:** - Given $m = \frac{13}{8}$, $$Q_m = \begin{bmatrix} 1 & \frac{13}{8} \\ 0 & 1 \end{bmatrix}$$ - Multiply: $$Q_m \mathbf{x} = \begin{bmatrix} 1 & \frac{13}{8} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{11}{7} \\ \frac{18}{11} \end{bmatrix} = \begin{bmatrix} \frac{11}{7} + \frac{13}{8} \cdot \frac{18}{11} \\ \frac{18}{11} \end{bmatrix}$$ - Simplify the first component: $$\frac{13}{8} \cdot \frac{18}{11} = \frac{13 \cdot 18}{8 \cdot 11} = \frac{234}{88} = \frac{117}{44}$$ $$x_1 = \frac{11}{7} + \frac{117}{44} = \frac{11 \cdot 44}{7 \cdot 44} + \frac{117 \cdot 7}{44 \cdot 7} = \frac{484}{308} + \frac{819}{308} = \frac{1303}{308}$$ 5. **Final answers:** $$R_\theta \mathbf{x} = \begin{bmatrix} -\frac{247 \sqrt{2}}{154} \\ -\frac{5 \sqrt{2}}{154} \end{bmatrix}$$ $$Q_m \mathbf{x} = \begin{bmatrix} \frac{1303}{308} \\ \frac{18}{11} \end{bmatrix}$$ These are the transformed vectors after rotation and shear respectively.