Example 11 - Prove |a a+b a+b+c 2a 3a+2b - Chapter 4 NCERT - Solving by simplifying det.

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Example 11 Prove that + + + 2 3 +2 4 +3 +2 3 6 +3 10 +6 +3 = a3 Let = + + + 2 3 +2 4 +3 +2 3 6 +3 10 +6 +3 Applying R2 R2 2R1 = + + + ( ) 3 +3 2( + ) 4 +3 +2 2( + + ) 3 6 +3 10 +6 +3 = + + + 3 +3 2 2 4 +3 +2 2 2 2 3 6 +3 10 +6 +3 = + + + 2 + 3 6 +3 10 +6 +3 Applying R3 R3 3R1 = + + + 0 2 + ( ) 6 +3 3( + ) 10 +6 +3 3 + + = + + + 0 2 + 6 +3 3 3 10 +6 +3 3 + + = + + + 0 2 + 0 3 7 +3 Expanding along C1 = = a 2 + 3 7 +3 0 + 0 = a (a(7a + 3b) 3a (2a + b) = a (7a2 + 3ab 6a2 3ab) = a(a2) = a3 = R.H.S Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.