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Ex 4.4, 4 - Using Cofactors of elements of third column, evaluate

Ex 4.4, 4 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.4, 4 - Chapter 4 Class 12 Determinants - Part 3

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Ex 4.4, 4 Using Cofactors of elements of third column, evaluate ∆ = |■8(1&x&yz@1&y&zx@1&z&xy)| ∆ = |■8(1&x&yz@1&y&zx@1&z&xy)| ∆ = a13 A13 + a23 A23 + a33 A33 a13 = yz , a23 = zx , a33 = xy , Calculating cofactors of third column i.e. A13 , A23 , And A33 M13 = |■8(1&x&yz@1&y&zx@1&z&xy)| = |■8(1&y@1&𝑧)| = 1 × z – 1 × y = z – y M23 = |■8(1&x&yz@1&y&zx@1&z&xy)| = |■8(1&x@1&z)| = 1 × z – 1 × x = z – x M33 = |■8(1&x&yz@1&y&zx@1&z&xy)| = |■8(1&x@1&𝑦)| = 1 × y – 1 × x = y – x A13 = (–1)1+3 M13 = (–1)4 . (z – y) = z – y A23 = (–1)2+3 . M23 = (–1)5 . (z – x) = (–1) (z – x) = x – z A33 = (–1)3 + 3 . M33 = (–1)6 . M33 = 1 . (y – x) = y – x Cofactor of 𝑎_𝑖𝑗= 𝐴_𝑖𝑗 = (−1)^(𝑖+𝑗). 𝑀_𝑖𝑗 Now, ∆ = a13 A13 + a23 A23 + a33 A33 = yz (z – y) + zx (x – z) + xy (y – x) = yz2 – y2z + zx2 – z2x + xy2 – x2y = (yz2 – y2z) + (xy2 – z2x) + (zx2 – x2y) = yz (z – y) + x (y2 – z2) + x2 (z – y) = – yz (y – z) + x (y2 – z2) – x2 (y – z) = – yz (y – z) + x (y + z) (y – z) – x2 (y – z) = (y – z) (−𝑦𝑧+𝑥(𝑦+𝑧)−𝑥2) = (y – z) (−𝑦𝑧+𝑥𝑦+𝑥𝑧−𝑥2) = (y – z) (𝑧(𝑥−𝑦)+𝑥(𝑦−𝑥)) = (y – z)(𝑧(𝑥−𝑦)−𝑥(𝑥−𝑦)) = (y – z)((𝑧−𝑥) (𝑥−𝑦)) = (x – y) (y – z) (z – 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.