$\begin{align}  & A{{A}^{-1}}=I \\ & \text{Since }\left( \begin{matrix}   -2 & 1  \\   2 & 3  \\\end{matrix} \right)\left( \begin{matrix}   p & q  \\   r & s  \\\end{matrix} \right)\text{gives}\left( \begin{matrix}   1 & 0  \\   0 & 1  \\\end{matrix} \right)\text{ which is an } \\ & \text{identity matrix, it means that }\left( \begin{matrix}   p & q  \\   r & s  \\\end{matrix} \right)\text{ is the inverse of}\left( \begin{matrix}   -2 & 1  \\   2 & 3  \\\end{matrix} \right) \\ & \text{For any given 2}\times \text{2 matrix, if }A=\left( \begin{matrix}   a & b  \\   c & d  \\\end{matrix} \right),\text{ then }{{A}^{-1}}=\frac{1}{ad-bc}\left( \begin{matrix}   d & -b  \\   -c & a  \\\end{matrix} \right) \\ & \text{Let }{{A}^{-1}}=\left( \begin{matrix}   -2 & 1  \\   2 & 3  \\\end{matrix} \right) \\ & {{A}^{-1}}=\frac{1}{(-2)(3)-(1)(2)}\left( \begin{matrix}   3 & -1  \\   -2 & -2  \\\end{matrix} \right) \\ & {{A}^{-1}}=-\frac{1}{8}\left( \begin{matrix}   3 & -1  \\   -2 & -2  \\\end{matrix} \right) \\ & {{A}^{-1}}=\left( \begin{matrix}   -\tfrac{3}{8} & \tfrac{1}{8}  \\   \tfrac{2}{8} & \tfrac{2}{8}  \\\end{matrix} \right)=\left( \begin{matrix}   -\tfrac{3}{8} & \tfrac{1}{8}  \\   \tfrac{1}{4} & \tfrac{1}{4}  \\\end{matrix} \right) \\ & \therefore r=\tfrac{1}{4} \\\end{align}$