Let A and B be two invertible matrices of order 3×3. If det⁡ABAT=8 and det⁡AB−1=8, then det⁡BA−1BT is equal to

Given, ABAT=8⇒A∥B∥AT=8 [∵|XY|=|X|Y∣]∴|A|2|B|=8 …(i)∵AT=|A| Also __, we have AB−1=8⇒A∥B−1=8⇒        |A||B|=8    ...(ii)∵|A−1|=|A|−1=1|A|On multiplying Eqs.(i) and (ii) , we get |A|3=8.8=43⇒|A|=4⇒|B|=|A|8=48=12 Now, BA−1BT=|B|1|A||B|=121412=116