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Q.

Let  x1,x2,x3,x4,x5>0 and  A=[aij]5×5 matrix such that aij={|xixj|ijxi,i=j  if  (x12x3x5)(x22x3x5)0,   (x22x4x1)(x32x4x1)0,  (x32x5x2)(x42x5x2)0,   (x42x1x3)(x52x1x3)0,
(x52x2x4)(x12x2x4)0,  then the sum of face value of the number det A is (given that x3=3 )
 

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answer is 9.

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Detailed Solution

Add all the inequalitites and multiply by 2,
 (x1x2x2x3)2+(x1x3x3x4)2+(x1x4x4x3)2+(x1x5x5x4)2+....+....+....+0
x1=x2=x3=x4=x5=3,detA=35=243 . Sum of the digits of det A = 2 + 4 + 3 = 9
 

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