**Ex 4.2, 12**

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 4.2, 12 By using properties of determinants, show that: 1xx2x21xxx21 = (1 – x3)2 Taking L.H.S 1xx2x21xxx21 Applying R1 → R1 + R2 + R3 = 𝟏+𝐱𝟐+𝐱𝐱+𝟏+𝐱𝟐𝐱𝟐+𝐱+𝟏x21xxx21 Taking (1 + x + x2) Common from 1st row = (𝟏+𝐱+𝐱𝟐) 111x21xxx21 Applying C1 → C1 − C2 = (1+x+x2) 𝟏−𝟏11x2−11xx−x2x21 = (1+x+x2) 011x2−11xx(1−x)x21 = (1+x+x2) 011(x−1)(𝐱−𝟏)1x−x (𝐱−𝟏)x21 Taking (x – 1) common from C1 = (x – 1) (1+x+x2) 011(x+1)1x−xx21 Applying C2 → C2 − C3 = (x – 1) (1+x+x2) 0𝟏−𝟏1x+11−xx−xx2−11 = (x – 1) (1+x+x2) 0𝟎1x+1−(𝒙−𝟏)x−x x−1(𝐱−𝟏)1 Taking (x – 1) common from 2nd Column = (x – 1) (1+x+x2) (x – 1) 001x+1−1x−x𝑥+11 Expanding Determinant along R1 = (x – 1)2 (1+x+x2) 0 −1𝑥𝑥+11−0 𝑥+1x−𝑥1+1 𝑥+1−1−x𝑥+1 = (x – 1)2 (1+x+x2) 0−0+1 𝑥+1−1−x𝑥+1 = (x – 1)2 (1+x+x2) 0−0+1( 𝑥+12−𝑥) = (x – 1)2 (1+x+x2) 𝑥+12−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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Davneet Singh

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