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))2 is
A
64
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B
-64
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C
512
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D
-512
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Solution
The correct option is D -512 (A3+B)A2−2(A3+B)C=0 (A3+B)(A2−2C)=0 →A3+B=0 (Since it is given that |2C−A2|=30, Inverse of (2C−A2) exists and is not a null matrix thus making A3+B=0) So B=−A3=−A(A2)=−A(−2I)=2A B =2A |B|=23|A| →|B|2=64|A|2 We know that A2=−2I |A|2=|−2I|⇒|A|2=−8×1=−8 Therefore |B|2=64|A|2=64×−8=−512