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Ex 4.2, 12 - Show |1 x x2 x2 1 x x x2 1| = (1 - x3)2 - Ex 4.2

Ex 4.2, 12 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.2, 12 - Chapter 4 Class 12 Determinants - Part 3 Ex 4.2, 12 - Chapter 4 Class 12 Determinants - Part 4 Ex 4.2, 12 - Chapter 4 Class 12 Determinants - Part 5

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Ex 4.2, 12 By using properties of determinants, show that: |■8(1&x&[email protected]&1&[email protected]&x2&1)| = (1 – x3)2 Solving L.H.S |■8(1&x&[email protected]&1&[email protected]&x2&1)| Applying R1 → R1 + R2 + R3 = |■8(𝟏+𝐱𝟐+𝐱&𝐱+𝟏+𝐱𝟐&𝐱𝟐+𝐱+𝟏@x2&1&[email protected]&x2&1)| Taking (1 + x + x2) Common from 1st row = (𝟏+𝐱+𝐱𝟐) |■8(1&1&[email protected]&1&[email protected]&x2&1)| Applying C1 → C1 − C2 = (1+x+x2) |■8(𝟏−𝟏&1&[email protected]−1&1&[email protected]−x2&x2&1)| = (1+x+x2) |■8(0&1&[email protected]−1&1&[email protected](1−x)&x2&1)| = (1+x+x2) |■8(𝟎&1&[email protected](𝐱−𝟏)(𝑥+1)&1&[email protected]−x (𝐱−𝟏)&x2&1)| Taking (x – 1) common from C1 = (x – 1) (1+x+x2) |■8(0&1&[email protected](x+1)&1&[email protected]−x&x2&1)| Applying C2 → C2 − C3 = (x – 1) (1+x+x2) |■8(0&𝟏−𝟏&[email protected]+1&1−x&[email protected]−x&x2−1&1)| = (x – 1) (1+x+x2) |■8(0&𝟎&[email protected]+1&−(𝒙−𝟏)&[email protected]−x&(x+1)(𝐱−𝟏)&1)| Taking (x – 1) common from 2nd Column = (x – 1) (1+x+x2) (x – 1) |■8(0&0&[email protected]+1&−1&[email protected]−x&𝑥+1&1)| = (x – 1)2 (1+x+x2) |■8(0&0&[email protected]+1&−1&[email protected]−x&𝑥+1&1)| Expanding Determinant along R1 = (x – 1)2 (1+x+x2) ( 0|■8(−1&𝑥@𝑥+1&1)|−0|■8(𝑥+1&[email protected]−𝑥&1)|+1|■8(𝑥+1&−[email protected]−x&𝑥+1)|) = (x – 1)2 (1+x+x2) ( 0−0+1|■8(𝑥+1&−[email protected]−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 = (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 We know that a3 – b3 = (a – b)(a2 + b2 + ab) Here, a = 1 , b = x

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.