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(B) Let $\Delta = \left| \begin{matrix} x - 4 & 2x & 2x \ 2x & x - 4 & 2x \ 2x & 2x & x - 4 \end{matrix} ight|$.Applying the operation $R_1 	o R_

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$(-4, 5)$

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(B) Let $\Delta = \left| \begin{matrix} x - 4 & 2x & 2x \ 2x & x - 4 & 2x \ 2x & 2x & x - 4 \end{matrix} ight|$.Applying the operation $R_1 	o R_1 + R_2 + R_3$,we get:$\Delta = \left| \begin{matrix} 5x - 4 & 5x - 4 & 5x - 4 \ 2x & x - 4 & 2x \ 2x & 2x & x - 4 \end{matrix} ight| = (5x - 4) \left| \begin{matrix} 1 & 1 & 1 \ 2x & x - 4 & 2x \ 2x & 2x & x - 4 \end{matrix} ight|$.Applying $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\Delta = (5x - 4) \left| \begin{matrix} 1 & 0 & 0 \ 2x & -x - 4 & 0 \ 2x & 0 & -x - 4 \end{matrix} ight| = (5x - 4)(-x - 4)^2 = (5x - 4)(x + 4)^2$.Comparing this with $(A + Bx)(x - A)^2$,we have $(A + Bx)(x - A)^2 = (5x - 4)(x - (-4))^2$.Thus,$A = -4$ and $B = 5$.Therefore,the ordered pair $(A, B) = (-4, 5)$.

If $\left| \begin{matrix} x - 4 & 2x & 2x \\ 2x & x - 4 & 2x \\ 2x & 2x & x - 4 \end{matrix} \right| = (A + Bx)(x - A)^2$,then the ordered pair $(A, B) = $ . . . . .

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