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Question

If A is a n×n non-singular matrix such that 3ABA1+A=2A1BA, then which of the following options is (are) CORRECT?

A
|A+B|=1
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B
|A+B|=0
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C
|ABA1A1BA|=0
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D
|ABA1A1BA|=1
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Solution

The correct options are
B |A+B|=0
C |ABA1A1BA|=0
3ABA1+A=2A1BA
Adding 2A both sides,
3ABA1+3A=2A1BA+2A
3A(BA1+I)=(A1B+I)2A
3A(BA1+AA1)=(A1B+A1A)2A
3A(B+A)A1=A1(B+A)2A
Taking determinant both sides,
|3A(B+A)A1|=|A1(B+A)2A|
3n|A||B+A||A1|=|A1||B+A|2n|A|
3n|B+A|=2n|B+A|
|B+A|=0


3ABA1+A=2A1BA
3ABA13A1BA=AA1BA
3(ABA1A1BA)=(A+A1BA)
3(ABA1A1BA)=(I+A1B)A
3(ABA1A1BA)=A1(A+B)A
Taking determinant both sides,
3n|ABA1A1BA|=(1)n|A1||A+B||A|
So, |ABA1A1BA|=0

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