Properties of Determinant

Chapter 4 Class 12 Determinants
Serial order wise

### Transcript

Question 12 By using properties of determinants, show that: |■8(1&x&x2@x2&1&x@x&x2&1)| = (1 – x3)2 Solving L.H.S |■8(1&x&x2@x2&1&x@x&x2&1)| Applying R1 → R1 + R2 + R3 = |■8(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) |■8(1&1&1@x2&1&x@x&x2&1)| Applying C1 → C1 − C2 = (1+x+x2) |■8(𝟏−𝟏&1&1@x2−1&1&x@x−x2&x2&1)| = (1+x+x2) |■8(0&1&1@x2−1&1&x@x(1−x)&x2&1)| = (1+x+x2) |■8(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) |■8(0&1&1@(x+1)&1&x@−x&x2&1)| Applying C2 → C2 − C3 = (x – 1) (1+x+x2) |■8(0&𝟏−𝟏&1@x+1&1−x&x@−x&x2−1&1)| = (x – 1) (1+x+x2) |■8(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) |■8(0&0&1@x+1&−1&x@−x&𝑥+1&1)| = (x – 1)2 (1+x+x2) |■8(0&0&1@x+1&−1&x@−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&x@−𝑥&1)|+1|■8(𝑥+1&−1@−x&𝑥+1)|) = (x – 1)2 (1+x+x2) ( 0−0+1|■8(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 = (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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.