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If A is a square matrix of order 3 and ∣adjA∣=49|adj A| = 49, then ∣7A−1∣2|7A^{-1}|^2

Solution

✅ Correct Option: 2

∣adjA∣=∣A∣n−1=∣A∣2=49|adj A| = |A|^{n-1} = |A|^2 = 49, so ∣A∣=7|A| = 7 (taking positive value).

∣7A−1∣=73⋅∣A−1∣=343∣A∣=3437=49|7A^{-1}| = 7^3 \cdot |A^{-1}| = \frac{343}{|A|} = \frac{343}{7} = 49.

∣7A−1∣2=492=74|7A^{-1}|^2 = 49^2 = 7^4.

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