Candidates preparing for JEE Advanced 2026 can download the official syllabus PDF from jeeadv.ac.in, released by IIT Roorkee.

Solution: det (A) = −5, and for n×n matrix adj (A) has determinant (det A)^ (n−1). Here n = 3, so det (B) = (−5)^ (2) = 25.

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