Slide45.JPG

Slide46.JPG
Slide47.JPG Slide48.JPG Slide49.JPG Slide50.JPG Slide51.JPG Slide52.JPG

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


Transcript

Ex 4.4, 12 Let A = [■8(3&7@2&5)] and B = [■8(6&8@7&9)] verify that (AB)-1 = B-1 A-1 Taking L.H.S (AB)–1 First calculating AB AB = [■8(3&7@2&5)] [■8(6&8@7&9)] = [■8(3(6)+7(7)&3(8)+7(9)@2(6)+5(7)&2(8)+5(9))] = [■8(18+49&24+63@12+35&16+45)] = [■8(67&87@47&61)] Now, (AB)-1 = 1/(|AB|) adj (AB) exists if |AB| ≠ 0 |AB| = |■8(67&87@47&61)| = 67 (61) – 47(87) = 4087 – 4089 = –2 Since |AB| ≠ 0 ∴ (AB)–1 exists Now, AB = [■8(67&87@47&61)] adj (AB) = [■8(67&87@47&61)] = [■8(61&−87@−47&67)] Thus, (AB)–1 = 1/(|AB|) adj (AB) Putting values = 1/(−2) [■8(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| = |■8(6&8@7&9)| = 6(9) – 7(8) = 54 – 56 = –2 Since |B|≠ 0 ∴ B–1 exists Now, B = [■8(6&8@7&9)] adj B = [■8(6&8@7&9)] = [■8(9&−8@−7&6)] Thus, B–1 = 1/(|B|) adj (B) = 1/(−2) [■8(9&−8@−7&6)] Calculating A-1 A-1 = 1/(|A|) adj (A) exist if |A| ≠ 0 |A| = |■8(3&7@2&5)| = 15 – 14 = 1 Since |A| ≠ 0, A-1 exists A = [■8(3&7@2&5)] adj A = [■8(3&7@2&5)] = [■8(5&−7@−2&3)] So, A–1 = 1/(|A|) adj (A) = 1/1 [■8(5&−7@−2&3)] = [■8(5&−7@−2&3)] Now B-1 A-1 = (−1)/2 [■8(9&−8@−7&6)] [■8(5&−7@−2&3)] = (−1)/2 [■8(9(5)+( –8)( –2)&9(−7)+(−8)(3)@ –7(5)+6( –2)&−7(−7)+6(3))] = (−1)/2 [■8(45+16&−63−24@−35−12&49+18)] = (−1)/2 [■8(61&−87@−47&67)] = L.H.S ∴ L.H.S = R.H.S Hence proved

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.