1. The problem asks to find the inverse relation $r^{-1}$ of the relation $r = \{ (x,y) \in \mathbb{R} \times \mathbb{R} \mid y = x^2 \}$. 2. The relation $r$ consists of all pairs $(x,y)$ where $y = x^2$. The inverse relation $r^{-1}$ swaps the coordinates, so $r^{-1} = \{ (y,x) \mid (x,y) \in r \}$. 3. Since $y = x^2$, swapping gives $x = y^2$ or equivalently $y = \pm \sqrt{x}$ when solving for $y$ in terms of $x$ in the inverse relation. 4. Therefore, the inverse relation $r^{-1}$ is $\{ (x,y) \in \mathbb{R} \times \mathbb{R} \mid y = \pm \sqrt{x} \}$, which matches option ก. 5. Options ข and ง only include the positive or negative square root, which is incomplete for the inverse relation. Option ค is incorrect because $y^2 = x^2$ does not represent the inverse relation of $y = x^2$. Final answer: ก. $r^{-1} = \{ (x,y) \in \mathbb{R} \times \mathbb{R} \mid y = \pm \sqrt{x} \}$