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Ex 4.2, 12 - Show |1 x x2 x2 1 x x x2 1| = (1 - x3)2 - Making whole row/column one and simplifying

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.2, 12 By using properties of determinants, show that: 1﷮x﷮x2﷮x2﷮1﷮x﷮x﷮x2﷮1﷯﷯ = (1 – x3)2 Taking L.H.S 1﷮x﷮x2﷮x2﷮1﷮x﷮x﷮x2﷮1﷯﷯ Applying R1 → R1 + R2 + R3 = 𝟏+𝐱𝟐+𝐱﷮𝐱+𝟏+𝐱𝟐﷮𝐱𝟐+𝐱+𝟏﷮x2﷮1﷮x﷮x﷮x2﷮1﷯﷯ Taking (1 + x + x2) Common from 1st row = (𝟏+𝐱+𝐱𝟐) 1﷮1﷮1﷮x2﷮1﷮x﷮x﷮x2﷮1﷯﷯ Applying C1 → C1 − C2 = (1+x+x2) 𝟏−𝟏﷮1﷮1﷮x2−1﷮1﷮x﷮x−x2﷮x2﷮1﷯﷯ = (1+x+x2) 0﷮1﷮1﷮x2−1﷮1﷮x﷮x(1−x)﷮x2﷮1﷯﷯ = (1+x+x2) 0﷮1﷮1﷮(x−1)(𝐱−𝟏)﷮1﷮x﷮−x (𝐱−𝟏)﷮x2﷮1﷯﷯ Taking (x – 1) common from C1 = (x – 1) (1+x+x2) 0﷮1﷮1﷮(x+1)﷮1﷮x﷮−x﷮x2﷮1﷯﷯ Applying C2 → C2 − C3 = (x – 1) (1+x+x2) 0﷮𝟏−𝟏﷮1﷮x+1﷮1−x﷮x﷮−x﷮x2−1﷮1﷯﷯ = (x – 1) (1+x+x2) 0﷮𝟎﷮1﷮x+1﷮−(𝒙−𝟏)﷮x﷮−x﷮ x−1﷯(𝐱−𝟏)﷮1﷯﷯ Taking (x – 1) common from 2nd Column = (x – 1) (1+x+x2) (x – 1) 0﷮0﷮1﷮x+1﷮−1﷮x﷮−x﷮𝑥+1﷮1﷯﷯ Expanding Determinant along R1 = (x – 1)2 (1+x+x2) 0 −1﷮𝑥﷮𝑥+1﷮1﷯﷯−0 𝑥+1﷮x﷮−𝑥﷮1﷯﷯+1 𝑥+1﷮−1﷮−x﷮𝑥+1﷯﷯﷯ = (x – 1)2 (1+x+x2) 0−0+1 𝑥+1﷮−1﷮−x﷮𝑥+1﷯﷯﷯ = (x – 1)2 (1+x+x2) 0−0+1( 𝑥+1﷯2−𝑥)﷯ = (x – 1)2 (1+x+x2) 𝑥+1﷯﷮2﷯−𝑥﷯ = (x – 1)2 (1 + x + x2) ((x2 + 1 + 2x) – x) = (x – 1)2 (1 + x + x2) (1 + x + x2) = (x – 1)2 (1 + x + x2)2 = ((x – 1) (1 + x + x2))2 = (– (1 – x) (1 + x + x2))2 = ((1 – x) (1 + x + x2))2 = (13 – x3)2 = (1 – x3)2 = R.H.S Hence proved

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