# Ex 4.2, 11

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 4.2, 11 By using properties of determinants, show that: (i) a b c 2a 2a 2b b c a 2b 2c 2c c a b = (a + b + c)3 Taking L.H.S = a b c 2a 2a 2b b c a 2b 2c 2c c a b Applying R1 R1 + R2 + R3 = a b c+2 +2 2a+b c a+2c 2a+2 + 2b b c a 2b 2c 2c c a b = + + + + + + 2b b c a 2b 2c 2c c a b Taking common (a + b + c) from R1 = (a+b+c) 1 1 1 2b 2 2c 2 c a b Applying C1 C1 C2 = (a+b+c) 1 1 2b (b c a) b c a 2 2c 2c 2 = (a+b+c) 1 1 b+c+a b c a 2 0 2 Applying C2 C2 C3 = ( + + ) 0 1 + + 2 2 0 2 ( ) = ( + + ) 0 1 + + 2 0 + + Taking (a + b + c) common from C1 = ( + + )( + + ) 0 0 1 1 2 0 ( + + ) = + + 2 0 0 1 1 ( + + ) 2 0 ( + + ) Taking common (a + b + c) from C2 = + + 2(a+b+c) 0 0 1 1 1 2 0 1 Expanding determinant along R1 = (a + b + c)3 0 1 2b 1 c a b 0 1 2b 0 c a b +1 1 1 0 1 = (a + b + c)3 0 0+1(1 0) = (a + b + c)3 1 = (a + b + c)3 Hence proved Ex 4.2, 11 By using properties of determinants, show that: (ii) x+y+2z x y z y+z+2x y z x z+x+2y = 2(x+y+z)3 Taking L.H.S x+y+2z x y z y+z+2x y z x z+x+2y Applying C1 C1 + C2 + C3 = + +2 + + z+y+z+2x+y y+ +2 y z+x+z+x+2y x z+x+2y = ( + + ) ( + + ) y+ +2 y ( + + ) x z+x+2y Taking common 2( + + ) from C1 = ( + + ) 1 1 y+ +2 y 1 x z+x+2y Applying R2 R2 R3 = 2 x+y+z 1 y+ +2 y ( + +2 ) 1 x z+x+2y = 2(x+y+z) 1 + + 1 x z+x+2y = 2(x+y+z) 1 0 ( + + ) ( + + ) 1 x z+x+2y Taking common ( + + ) from 2nd Row = 2(x+y+z)(x+y+z) 1 0 1 1 1 x z+x+2y Applying R3 R3 R1 = 2 x+y+z 2 1 0 1 1 x z+x+2y y = 2 x+y+z 2 1 0 1 1 0 x+y+z Taking common ( + + ) Common from 3rd Row = 2 x+y+z 2 x+y+z 1 0 1 1 0 0 1 Expanding Determinant along C1 = 2 x+y+z 3 1 1 1 0 1 0 0 1 +0 x y 1 1 = 2 x+y+z 3 1 1 1 0 1 0+0 = 2 x+y+z 3 1 1 0 0 + (0) = 2 x+y+z 3 1 = 2 x+y+z 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 9 years. He provides courses for Maths and Science at Teachoo.