# Ex 4.2, 11

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 4.2, 11 By using properties of determinants, show that: (i) a−b−c2a2a2bb−c−a2b2c2cc−a−b = (a + b + c)3 Taking L.H.S = a−b−c2a2a2bb−c−a2b2c2cc−a−b Applying R1 → R1 + R2 + R3 = a−b−c+2𝑏+2𝑐2a+b−c−a+2c2a+2𝑏+𝑐−𝑎−𝑏2bb−c−a2b2c2cc−a−b = 𝐚+𝐛+𝐜𝐚+𝐛+𝐜𝐚+𝒃+𝒄2bb−c−a2b2c2cc−a−b Taking common (a + b + c) from R1 = (a+b+c) 1112b𝑏−𝑐−𝑎2𝑏2c2𝑐c−a−b Applying C1 → C1 – C2 = (a+b+c) 𝟏−𝟏112b−(b−c−a)b−c−a2𝑏2c−2c2𝑐𝑐−𝑎−𝑏 = (a+b+c) 𝟎11b+c+ab−c−a2𝑏02𝑐𝑐−𝑎−𝑏 Applying C2 → C2 – C3 = (𝑎+𝑏+𝑐) 0𝟏−𝟏1𝑏+𝑐+𝑎𝑏−𝑐−𝑎−2𝑏2𝑏02𝑐−(𝑐−𝑎−𝑏)𝑐−𝑎−𝑏 = (𝑎+𝑏+𝑐) 0𝟎1𝑏+𝑐+𝑎−𝑏−𝑐−𝑎2𝑏0𝑎+𝑏+𝑐𝑐−𝑎−𝑏 Taking (a + b + c) common from C1 = (𝑎+𝑏+𝑐)(𝑎+𝑏+𝑐) 0011−𝑏−𝑐−𝑎2𝑏0(𝑎+𝑏+𝑐)𝑐−𝑎−𝑏 = 𝑎+𝑏+𝑐2 0011−(𝒂+𝒃+𝒄)2𝑏0(𝒂+𝒃+𝒄)𝑐−𝑎−𝑏 Taking common (a + b + c) from C2 = 𝑎+𝑏+𝑐2(a+b+c) 0011−12𝑏01𝑐−𝑎−𝑏 Expanding determinant along R1 = (a + b + c)3 0 −12b1c−a−b−0 12b0c−a−b+1 1−101 = (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+2zxyzy+z+2xyzxz+x+2y = 2(x+y+z)3 Taking L.H.S x+y+2zxyzy+z+2xyzxz+x+2y Applying C1 → C1 + C2 + C3 = 𝑥+𝑦+2𝑧+𝑥+𝑦𝑥𝑦z+y+z+2x+yy+𝑧+2𝑥yz+x+z+x+2yxz+x+2y = 𝟐(𝒙+𝒚+𝒛)𝑥𝑦𝟐(𝒙+𝒚+𝒛)y+𝑧+2𝑥y𝟐(𝒙+𝒚+𝒛)xz+x+2y Taking common 2(𝑥+𝑦+𝑧) from C1 = 𝟐(𝐱+𝐲+𝐳) 1𝑥𝑦1y+𝑧+2𝑥y1xz+x+2y Applying R2 → R2 – R3 = 2 x+y+z 1𝑥𝑦𝟏−𝟏y+𝑧+2𝑥−𝑥y−(𝑧+𝑥+2𝑦)1xz+x+2y = 2(x+y+z) 1𝑥𝑦𝟎𝑥+𝑦+𝑧−𝑥−𝑦−𝑧1xz+x+2y = 2(x+y+z) 1𝑥𝑦0(𝒙+𝒚+𝒛)−(𝒙+𝒚+𝒛)1xz+x+2y Taking common (𝑥+𝑦+𝑧) from 2nd Row = 2(x+y+z)(x+y+z) 1𝑥𝑦01−11xz+x+2y Applying R3 → R3 – R1 = 2 x+y+z2 1𝑥𝑦01−1𝟏−𝟏x−𝑥z+x+2y−y = 2 x+y+z2 1𝑥𝑦01−1𝟎0x+y+z Taking common (𝑥+𝑦+𝑧) Common from 3rd Row = 2 x+y+z2 x+y+z 1𝑥𝑦01−1001 Expanding Determinant along C1 = 2 x+y+z3 1 1−101−0 𝑥𝑦01+0 xy1−1 = 2 x+y+z3 1 1−101−0+0 = 2 x+y+z3 1 1−0−𝑥 0+𝑦(0) = 2 x+y+z3 1 = 2 x+y+z3 = R.H.S Hence proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.