Example 5 - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Examples
Example 1
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Chapter 4 Class 12 Determinants
Serial order wise
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
