x2+xx+1x−22x2+3x−13x3x−3x2+2x+32x−12x−1=xA+B, where A and B are real constants not involving x, then 3A=

R2→R2−2R1;R3→R3−R1x2+xx+1x−2x−1x−2x+1x+3x−2x+1R3→R3−R2x2+xx+1x−2x−1x−2x+1400=24x−12∴ A=24

x2+xx+1x−22x2+3x−13x3x−3x2+2x+32x−12x−1=xA+B, where A and B are real constants not involving x, then 3A=

R2→R2−2R1;R3→R3−R1x2+xx+1x−2x−1x−2x+1x+3x−2x+1R3→R3−R2x2+xx+1x−2x−1x−2x+1400=24x−12∴ A=24

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$|\begin{matrix} x^{2} + x & x + 1 & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix}| = x A + B,$ where A and B are real constants not involving $x,$ then $3 A =$

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answer is B.

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$R_{2} \to R_{2} - 2 R_{1}; R_{3} \to R_{3} - R_{1}$

$|\begin{matrix} x^{2} + x & x + 1 & x - 2 \\ x - 1 & x - 2 & x + 1 \\ x + 3 & x - 2 & x + 1 \end{matrix}|$

$R_{3} \to R_{3} - R_{2}$

$|\begin{matrix} x^{2} + x & x + 1 & x - 2 \\ x - 1 & x - 2 & x + 1 \\ 4 & 0 & 0 \end{matrix}| = 24 x - 12$

$∴ A = 24$

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x2+xx+1x−22x2+3x−13x3x−3x2+2x+32x−12x−1=xA+B, where A and B are real constants not involving x, then 3A=

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$|\begin{matrix} x^{2} + x & x + 1 & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix}| = x A + B,$ where A and B are real constants not involving $x,$ then $3 A =$