Let A,B,C be three square matrices of order 3×3 such that A2+2I=0, |(2C−A2)|=30 and satisfying the equation A5−2A3C+BA2−2BC=0
Value of (det.(B−A))2 is
A
8
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B
-8
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C
64
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D
-64
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Solution
The correct option is B -8 (A3+B)A2−2(A3+B)C=0 (A3+B)(A2−2C)=0 Since |2C−A2|=30, so inverse of (2C−A2) exists. So A3+B=0 B=−A3=−A(A2)=−A(−2I)=2A B =2A |B−A|2=|2A−A|2=|A|2 We know that A2=−2I |A|2=|−2I|⇒|A|2=−8×1=−8 |B|2=64|A|2=64×−8=−512 Therefore |B−A|2=|A|2=−8