1. We are given the matrix equation: $$\begin{pmatrix} x & s - x^2 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} x - 2 & 1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 9 \\ 6 & c \end{pmatrix}$$ where $x \in S$ and we need to find the values of $a$, $b$, and $c$ (assuming $a, b$ come from the entries of the result matrix or missing context). 2. Multiply the two matrices on the left: $$\begin{pmatrix} x & s - x^2 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} x - 2 & 1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} x(x - 2) + (s - x^2) \cdot 2 & x \cdot 1 + (s - x^2) \cdot 1 \\ -3(x - 2) + 1 \cdot 2 & -3 \cdot 1 + 1 \cdot 1 \end{pmatrix}$$ 3. Simplify each element: - Top left: $$x(x - 2) + 2(s - x^2) = x^2 - 2x + 2s - 2x^2 = -x^2 - 2x + 2s$$ - Top right: $$x + s - x^2$$ - Bottom left: $$-3(x - 2) + 2 = -3x + 6 + 2 = -3x + 8$$ - Bottom right: $$-3 + 1 = -2$$ 4. Set each element equal to the corresponding element of the right matrix: \begin{cases} -x^2 - 2x + 2s = 2 \\ x + s - x^2 = 9 \\ -3x + 8 = 6 \\ -2 = c \end{cases} 5. From the third equation: $$-3x + 8 = 6 \implies -3x = -2 \implies x = \frac{2}{3}$$ 6. Substitute $x = \frac{2}{3}$ into the second equation: $$\frac{2}{3} + s - \left(\frac{2}{3}\right)^2 = 9$$ $$\frac{2}{3} + s - \frac{4}{9} = 9$$ $$s + \frac{2}{3} - \frac{4}{9} = 9$$ Find common denominator $9$: $$s + \frac{6}{9} - \frac{4}{9} = 9 \implies s + \frac{2}{9} = 9$$ $$s = 9 - \frac{2}{9} = \frac{81}{9} - \frac{2}{9} = \frac{79}{9}$$ 7. Substitute $x = \frac{2}{3}$ and $s = \frac{79}{9}$ into the first equation: $$-\left(\frac{2}{3}\right)^2 - 2\left(\frac{2}{3}\right) + 2 \left(\frac{79}{9}\right) = 2$$ Calculate step-by-step: $$- \frac{4}{9} - \frac{4}{3} + \frac{158}{9} = 2$$ Convert all to ninths: $$- \frac{4}{9} - \frac{12}{9} + \frac{158}{9} = 2$$ $$\frac{-4 - 12 + 158}{9} = 2 \implies \frac{142}{9} = 2$$ Multiply both sides by 9: $$142 = 18$$ This is a contradiction, so the initial assumed variables or relations must be reconsidered. 8. Since the problem asks to find values of $a, b, c$ but only $c$ is present in the matrix, we interpret $a$ and $b$ as the unknowns in $s = a$ and $x = b$ or require clarification. 9. Given the contradiction, the only consistent values found are: $$x = \frac{2}{3}, \quad c = -2$$ Summary: $$x = \frac{2}{3},\quad c = -2,\quad s = \frac{79}{9}$$ "a" and "b" not explicitly defined; assuming $a = x$, $b = s$, answer in matrix form: $$\begin{pmatrix} x & s - x^2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & \frac{79}{9} - \left(\frac{2}{3}\right)^2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & \frac{79}{9} - \frac{4}{9} \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & \frac{75}{9} \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & \frac{25}{3} \\ -3 & 1 \end{pmatrix}$$