Example 5 - Chapter 4 Class 12 Determinants

Last updated at April 16, 2024 by Teachoo

Transcript

Example 5 Find values of x for which |■8(3&x@x&1)| = |■8(3&2@4&1)| Given |■8(3&x@x&1)| = |■8(3&2@4&1)| Putting values 3 – x2 = – 5 – x2 = – 5 – 3 – x2 = – 8 x2 = 8 Calculating |■8(3&x@x&1)| = 3(1) – x (x) = 3 – x2 Calculating |■8(3&2@4&1)| = 3(1) – 4(2) = 3 – 8 = – 5 x = ± √8 = ± √(2" ×" 2" ×" 2) = ± 2√2 Hence value of x is ± 2√𝟐

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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