1. Problem: Find $(RS)^{-1}$ given matrices $$R=\begin{bmatrix}3 & -5 \\ 3 & 9\end{bmatrix}, \quad S^{-1}=\begin{bmatrix}-1 & -2 \\ 5 & 3\end{bmatrix}$$ 2. Formula: The inverse of a product of matrices satisfies $$ (RS)^{-1} = S^{-1} R^{-1} $$ 3. We already have $S^{-1}$, so we need to find $R^{-1}$. 4. Calculate $\det(R)$: $$ \det(R) = (3)(9) - (3)(-5) = 27 + 15 = 42 $$ 5. Find $R^{-1}$ using the formula for 2x2 inverse: $$ R^{-1} = \frac{1}{\det(R)} \begin{bmatrix} 9 & 5 \\ -3 & 3 \end{bmatrix} = \frac{1}{42} \begin{bmatrix} 9 & 5 \\ -3 & 3 \end{bmatrix} $$ 6. Now compute $(RS)^{-1} = S^{-1} R^{-1}$: $$ S^{-1} R^{-1} = \begin{bmatrix} -1 & -2 \\ 5 & 3 \end{bmatrix} \times \frac{1}{42} \begin{bmatrix} 9 & 5 \\ -3 & 3 \end{bmatrix} = \frac{1}{42} \begin{bmatrix} (-1)(9) + (-2)(-3) & (-1)(5) + (-2)(3) \\ 5(9) + 3(-3) & 5(5) + 3(3) \end{bmatrix} $$ 7. Simplify the matrix multiplication: $$ = \frac{1}{42} \begin{bmatrix} -9 + 6 & -5 - 6 \\ 45 - 9 & 25 + 9 \end{bmatrix} = \frac{1}{42} \begin{bmatrix} -3 & -11 \\ 36 & 34 \end{bmatrix} $$ 8. Final answer: $$ (RS)^{-1} = \begin{bmatrix} -\frac{3}{42} & -\frac{11}{42} \\ \frac{36}{42} & \frac{34}{42} \end{bmatrix} = \begin{bmatrix} -\frac{1}{14} & -\frac{11}{42} \\ \frac{6}{7} & \frac{17}{21} \end{bmatrix} $$