Ex 4.2, 13 - Chapter 4 Class 12 Determinants
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 4.2, 13 By using properties of determinants, show that: 1+a2 b2 2ab 2b 2ab 1 a2+b2 2a 2b 2a 1 a2 b2 = (1 + a2+b2)3 Taking L.H.S 1+a2 b2 2ab 2b 2ab 1 a2+b2 2a 2b 2a 1 a2 b2 Applying R1 R1 + bR3 = 1+a2 b2+ (2 ) 2ab+b( 2a) 2b+ (1 2 2) 2ab 1 a2 b2 2a 2b 2a 1 a2 b2 = 1+a2 b2+2 2 2ab 2ab 2b+ 2 3 2ab 1 a2 b2 2a 2b 2a 1 a2 b2 = + + 0 ( + + ) 2ab 1 a2+b2 2a 2b 2a 1 a2 b2 Taking Common (1+ 2+ 2) from R1 = (1+ 2+ 2) 1 0 b 2ab 1 a2+b2 2a 2b 2a 1 a2 b2 Applying R2 R2 aR3 = (1+ 2+ 2) 1 0 b 2ab (2 ) 1 a2+b2 ( 2 ) 2a a(1 a2 b2) 2b 2a 1 a2 b2 = (1+ 2+ 2) 1 0 b 2ab 2 1 a2+b2+2 2 2a a+a3+ab2 2b 2a 1 a2 b2 = (1+ 2+ 2) 1 0 b 0 1+b2+ 2 a+a3+ab2 2b 2a 1 a2 b2 Taking Common (1+ 2+ 2) from R2 = (1+ 2+ 2)2 1 0 b 0 1 a 2b 2a 1 a2 b2 = (1+ 2+ 2)2 1 1 2 1 a2 b2 0 0 x 2 1 a2 b2 + 0 1 2b 2 = (1+ 2+ 2)2 1 1 2 1 a2 b2 0+ 0 1 2b 2 = (1+ 2+ 2)2 (1(1 a2 b2) + 2a2) b (0 2b)) = (1+ 2+ 2)2 (1 a2 b2 + 2a2 + 2a2) = (1+ 2+ 2)2 (1 + a2 + b2) = 1+ 2+ 2 3 = R.H.S Hence proved
Chapter 4 Class 12 Determinants
Serial order wise
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Davneet Singh
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.
