If A and B are two matrices of order '3' such that 3A+4BBT=I and Bā1=AT, then identify which of the following statements is/are correct?
A
Tr(A−1−4B3−BA+I)=9
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B
Tr(A−1−4B3+BA)=12
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C
|A2−3A3|=64
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D
|A−1−4B3−BA|=8
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Solution
The correct option is D|A−1−4B3−BA|=8 3A+4BBT=I ⇒3A=I−4BBT ⇒3AT=IT−4(BBT)T ⇒3AT=I−4BBT=3A
So A is symmetric.
So B−1=AT=A
So B−1=A⇒B=A−1 3A+4B.B=I3A+4(A−1)2=I3B−1+4B2=I3A2+4A−1=A3I+4B3=B=A−13A3+4I=A2A−1−4B3=3IA2−3A3=4I
So Tr(A−1−4B3−BA+I)=Tr(3I−I+I)=9 Tr(A−1−4B3+BA)=Tr(3I+I)=12 |A2−3A3|=|4I|=64 |A−1−4B3−BA|=|3I−I|=|2I|=8