1. Chapter 4 Class 12 Determinants
2. Serial order wise

Transcript

Ex 4.5, 12 Let A = 3﷮7﷮2﷮5﷯﷯ and B = 6﷮8﷮7﷮9﷯﷯ verify that (AB)-1 = B-1 A-1 Taking L.H.S (AB) –1 First calculating AB AB = 3﷮7﷮2﷮5﷯﷯ 6﷮8﷮7﷮9﷯﷯ = 3 6﷯+7(7)﷮3 8﷯+7(9)﷮2 6﷯+5(7)﷮2 8﷯+5(9)﷯﷯ = 18+49﷮24+63﷮12+35﷮16+45﷯﷯ = 67﷮87﷮47﷮61﷯﷯ Now, (AB)-1 = 1﷮|AB|﷯ adj (AB) exists if |AB| ≠ 0 |AB| = 67﷮87﷮47﷮61﷯﷯ = 67 (61) – 47(87) = 4087 – 4089 = – 2 Since |AB| ≠ 0 (AB) –1 exists AB = 67﷮87﷮47﷮61﷯﷯ adj (AB) = 67﷮87﷮47﷮61﷯﷯ = 61﷮−87﷮−47﷮67﷯﷯ Now, (AB)–1 = 1﷮|AB|﷯ adj (AB) Putting values = 1﷮−2﷯ 61﷮−87﷮−47﷮67﷯﷯ Taking R.H.S B-1A-1 First Calculating B-1 B–1 = 1﷮|B|﷯ adj (B) exist if |B|≠ 0 Now, |B| = 6﷮8﷮7﷮9﷯﷯ = 6(9) – 7(8) = 63 – 56 = – 2 B = 6﷮8﷮7﷮9﷯﷯ adj B = 6﷮8﷮7﷮9﷯﷯ = 9﷮−8﷮−7﷮6﷯﷯ Now, B–1 = 1﷮|B|﷯ adj (B) = 1﷮−2﷯ 9﷮−8﷮−7﷮6﷯﷯ Calculating A-1 A-1 = 1﷮|A|﷯ adj (A) exist if |A| ≠ 0 A﷯ = 3﷮7﷮2﷮5﷯﷯ = 15 – 14 = 1 Since |A| ≠ 0, A-1 exist A = 3﷮7﷮2﷮5﷯﷯ adj A = 3﷮7﷮2﷮5﷯﷯ = 5﷮−7﷮−2﷮3﷯﷯ So, A–1 = 1﷮|A|﷯ adj (A) = 1﷮1﷯ 5﷮−7﷮−2﷮3﷯﷯ = 5﷮−7﷮−2﷮3﷯﷯ Now B-1 A-1 = 1﷮2﷯ 9﷮−8﷮−7﷮6﷯﷯ 5﷮−7﷮−2﷮3﷯﷯ = 1﷮2﷯ 9 5﷯+( –8)( –2)﷮9 −7﷯+(−8)(3)﷮ –7 5﷯+6( –2)﷮−7 −7﷯+6(3)﷯﷯ = 1﷮2﷯ 45+16﷮−63−24﷮−35−12﷮49+18﷯﷯ = 1﷮2﷯ 61﷮−87﷮−47﷮67﷯﷯ = L.H.S ∴ L.H.S = R.H.S Hence proved

Serial order wise