# Ex 4.2, 7 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 4.2, 7 By using properties of determinants, show that: −a2abacba−b2bccacb−c2 = 4a2b2c2 Taking L.H.S −a2abacba−b2bccacb−c2 Taking a common from R1, b common from R2 , c common from R3 = abc −abca−bcab−c Taking a common from C1, b common from C2 , c common from C3 = abc (abc) −1111−1111−1 Applying C2 → C2 + C1 = (abc)2 −11−111−1+1111+1−1 = (abc)2 −10110112−1 Applying C3 → C3 + C1 = (abc)2 −101−1101+112−1+1 = (abc)2 −100102120 = (abc)2 −1 0220−0 1210+0 1012 = (abc)2 −1 0220−0+0 = (abc)2 ( – 1(0(0) – 2(2))) = (abc)2 (4) = 4 (abc)2 = 4a2b2c2 = R.H.S Hence proved

Ex 4.2

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 9 years. He provides courses for Maths and Science at Teachoo.