If a2+2a−52a+412a−4a+51−261>0, then

Δ=a2+2a−32a−202a−2a−10−261R1−R3;R2−R3=a2+2a−3(a−1)−(2a−2)2=(a+3)(a−1)2−4(a−1)2=(a−1)3>0 if a>1

If a2+2a−52a+412a−4a+51−261>0, then

Δ=a2+2a−32a−202a−2a−10−261R1−R3;R2−R3=a2+2a−3(a−1)−(2a−2)2=(a+3)(a−1)2−4(a−1)2=(a−1)3>0 if a>1

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If $|\begin{array}{ccc} a^{2} + 2 a - 5 & 2 a + 4 & 1 \\ 2 a - 4 & a + 5 & 1 \\ - 2 & 6 & 1 \end{array}| > 0$ , then

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a = 1

a < 1

a >1

a= 0

answer is A.

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Detailed Solution

$\begin{array}{l} Δ = |\begin{array}{ccc} a^{2} + 2 a - 3 & 2 a - 2 & 0 \\ 2 a - 2 & a - 1 & 0 \\ - 2 & 6 & 1 \end{array}| R_{1} - R_{3}; R_{2} - R_{3} \\ = (a^{2} + 2 a - 3) (a - 1) - (2 a - 2)^{2} \\ = (a + 3) (a - 1)^{2} - 4 (a - 1)^{2} = (a - 1)^{3} \\ > 0 if a > 1 \end{array}$