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If A and B are invertible matrices of order 3, ∣A∣=2|A| = 2 and ∣(AB)−1∣=−16|(AB)^{-1}| = -\frac{1}{6}, then the value of ∣B∣|B| is :

Solution

✅ Correct Option: 2

∣(AB)−1∣=1∣AB∣=1∣A∣∣B∣|(AB)^{-1}| = \frac{1}{|AB|} = \frac{1}{|A||B|}. So 12∣B∣=−16\frac{1}{2|B|} = -\frac{1}{6}, giving ∣B∣=−3|B| = -3.

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