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We are given the equation $PQ - RS = O$, where $O$ is the null matrix. This implies $PQ = RS$. To find $S$, we need to isolate it by multiplying both sides by the inverse of $R$ from the left: $R^{-1}PQ = S$.
Multiply matrices $P$ and $Q$: $$PQ = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} = \begin{bmatrix} (2)(5)+(-1)(7) & (2)(2)+(-1)(4) \\ (3)(5)+(4)(7) & (3)(2)+(4)(4) \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix}$$
For $R = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}$, the determinant $|R| = (2)(8) - (5)(3) = 16 - 15 = 1$. The inverse $R^{-1} = \frac{1}{|R|} \text{adj}(R) = \frac{1}{1} \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix}$.
Now compute $S = R^{-1}(PQ)$: $$S = \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} = \begin{bmatrix} (8)(3)+(-5)(43) & (8)(0)+(-5)(22) \\ (-3)(3)+(2)(43) & (-3)(0)+(2)(22) \end{bmatrix}$$ $$S = \begin{bmatrix} 24 - 215 & 0 - 110 \\ -9 + 86 & 0 + 44 \end{bmatrix} = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}$$
Final Answer: S = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}
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