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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To be eligible for JEE Advanced 2026, candidates must rank among the top 2.5 lakh students in JEE Main 2026, satisfy the age criteria. Additionally, candidates must have appeared for their Class 12 ...