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Question

Let A=4022x430x2B=3b1C=[263]
then the number of integral value (s) 'b' for which Tr(ABC)18xR is/are


Solution

ABC=2884422bx+46bx+123bx+6182x2546x2273x2)
Tr(ABC)=28+6bx+12273x218
3x26bx+250
D0
36b24×3×250
b2253253b253
b =-2,-1,0,1,2 are posible.

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