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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 4.3, 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 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.