Let A and B be two square matrices of order 3 such that det.(A)=3 and det. (B)=2, then the value of det. adj⋅B−1A−1−1 is

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answer is 36.

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Detailed Solution

adj⋅B−1A−1−1=adj⋅(AB)−1−1=(adj⋅(AB))−1−1=adj⋅(AB)∴det⁡adj⋅B−1A−1−1=|(adj⋅(AB))|=|AB|2=A2|B|2=9×4=36

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