Ex 4.2, 11

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+2𝑏+2𝑐﷮2a+b−c−a+2c﷮2a+2𝑏+𝑐−𝑎−𝑏﷮2b﷮b−c−a﷮2b﷮2c﷮2c﷮c−a−b﷯﷯ = 𝐚+𝐛+𝐜﷮𝐚+𝐛+𝐜﷮𝐚+𝒃+𝒄﷮2b﷮b−c−a﷮2b﷮2c﷮2c﷮c−a−b﷯﷯ Taking common (a + b + c) from R1 = (a+b+c) 1﷮1﷮1﷮2b﷮𝑏−𝑐−𝑎﷮2𝑏﷮2c﷮2𝑐﷮c−a−b﷯﷯ Applying C1 → C1 – C2 = (a+b+c) 𝟏−𝟏﷮1﷮1﷮2b−(b−c−a)﷮b−c−a﷮2𝑏﷮2c−2c﷮2𝑐﷮𝑐−𝑎−𝑏﷯﷯ = (a+b+c) 𝟎﷮1﷮1﷮b+c+a﷮b−c−a﷮2𝑏﷮0﷮2𝑐﷮𝑐−𝑎−𝑏﷯﷯ Applying C2 → C2 – C3 = (𝑎+𝑏+𝑐) 0﷮𝟏−𝟏﷮1﷮𝑏+𝑐+𝑎﷮𝑏−𝑐−𝑎−2𝑏﷮2𝑏﷮0﷮2𝑐−(𝑐−𝑎−𝑏)﷮𝑐−𝑎−𝑏﷯﷯ = (𝑎+𝑏+𝑐) 0﷮𝟎﷮1﷮𝑏+𝑐+𝑎﷮−𝑏−𝑐−𝑎﷮2𝑏﷮0﷮𝑎+𝑏+𝑐﷮𝑐−𝑎−𝑏﷯﷯ Taking (a + b + c) common from C1 = (𝑎+𝑏+𝑐)(𝑎+𝑏+𝑐) 0﷮0﷮1﷮1﷮−𝑏−𝑐−𝑎﷮2𝑏﷮0﷮(𝑎+𝑏+𝑐)﷮𝑐−𝑎−𝑏﷯﷯ = 𝑎+𝑏+𝑐﷯2 0﷮0﷮1﷮1﷮−(𝒂+𝒃+𝒄)﷮2𝑏﷮0﷮(𝒂+𝒃+𝒄)﷮𝑐−𝑎−𝑏﷯﷯ Taking common (a + b + c) from C2 = 𝑎+𝑏+𝑐﷯2(a+b+c) 0﷮0﷮1﷮1﷮−1﷮2𝑏﷮0﷮1﷮𝑐−𝑎−𝑏﷯﷯ 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 = 𝑥+𝑦+2𝑧+𝑥+𝑦﷮𝑥﷮𝑦﷮z+y+z+2x+y﷮y+𝑧+2𝑥﷮y﷮z+x+z+x+2y﷮x﷮z+x+2y﷯﷯ = 𝟐(𝒙+𝒚+𝒛)﷮𝑥﷮𝑦﷮𝟐(𝒙+𝒚+𝒛)﷮y+𝑧+2𝑥﷮y﷮𝟐(𝒙+𝒚+𝒛)﷮x﷮z+x+2y﷯﷯ Taking common 2(𝑥+𝑦+𝑧) from C1 = 𝟐(𝐱+𝐲+𝐳) 1﷮𝑥﷮𝑦﷮1﷮y+𝑧+2𝑥﷮y﷮1﷮x﷮z+x+2y﷯﷯ Applying R2 → R2 – R3 = 2 x+y+z﷯ 1﷮𝑥﷮𝑦﷮𝟏−𝟏﷮y+𝑧+2𝑥−𝑥﷮y−(𝑧+𝑥+2𝑦)﷮1﷮x﷮z+x+2y﷯﷯ = 2(x+y+z) 1﷮𝑥﷮𝑦﷮𝟎﷮𝑥+𝑦+𝑧﷮−𝑥−𝑦−𝑧﷮1﷮x﷮z+x+2y﷯﷯ = 2(x+y+z) 1﷮𝑥﷮𝑦﷮0﷮(𝒙+𝒚+𝒛)﷮−(𝒙+𝒚+𝒛)﷮1﷮x﷮z+x+2y﷯﷯ Taking common (𝑥+𝑦+𝑧) from 2nd Row = 2(x+y+z)(x+y+z) 1﷮𝑥﷮𝑦﷮0﷮1﷮−1﷮1﷮x﷮z+x+2y﷯﷯ Applying R3 → R3 – R1 = 2 x+y+z﷯2 1﷮𝑥﷮𝑦﷮0﷮1﷮−1﷮𝟏−𝟏﷮x−𝑥﷮z+x+2y−y﷯﷯ = 2 x+y+z﷯2 1﷮𝑥﷮𝑦﷮0﷮1﷮−1﷮𝟎﷮0﷮x+y+z﷯﷯ Taking common (𝑥+𝑦+𝑧) Common from 3rd Row = 2 x+y+z﷯2 x+y+z﷯ 1﷮𝑥﷮𝑦﷮0﷮1﷮−1﷮0﷮0﷮1﷯﷯ Expanding Determinant along C1 = 2 x+y+z﷯3 1 1﷮−1﷮0﷮1﷯﷯−0 𝑥﷮𝑦﷮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﷯−𝑥 0﷯+𝑦(0)﷯ = 2 x+y+z﷯3 1﷯ = 2 x+y+z﷯3 = R.H.S Hence proved