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Question

Let A,B,C be three square matrices of order 3×3 such that A2+2I=0, |(2CA2)|=30 and satisfying the equation A52A3C+BA22BC=0

Value of (det.(BA))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)A22(A3+B)C=0
(A3+B)(A22C)=0
Since |2CA2|=30, so inverse of (2CA2) exists.
So A3+B=0
B=A3=A(A2)=A(2I)=2A
B =2A
|BA|2=|2AA|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 |BA|2=|A|2=8

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