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.

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

The Joint Entrance Examination (JEE) Main 2025 Session 2 began on April 2, with Paper 1 (B.E./B.Tech) being conducted by the National Testing Agency (NTA). The first shift took place from 9:00 AM to ...

The JEE Advanced 2026 syllabus PDF is now available on the official website, jeeadv.ac.in. The detailed syllabus provides ...

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