   1. Chapter 4 Class 12 Determinants
2. Serial order wise
3. Ex 4.4

Transcript

Ex4.4, 4 Using Cofactors of elements of third column, evaluate ∆ = 1﷮x﷮yz﷮1﷮y﷮zx﷮1﷮z﷮xy﷯﷯ ∆ = 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 = 1﷮x﷮yz﷮1﷮y﷮zx﷮1﷮z﷮xy﷯﷯ = 1﷮y﷮1﷮𝑧﷯﷯ = 1 × z – 1 × y = z – y M23 = 1﷮x﷮yz﷮1﷮y﷮zx﷮1﷮z﷮xy﷯﷯ = 1﷮x﷮1﷮z﷯﷯ = 1 × z – 1 × x = z – x M33 = 1﷮x﷮yz﷮1﷮y﷮zx﷮1﷮z﷮xy﷯﷯ = 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)

Ex 4.4 