Check sibling questions

Ex 4.5, 12 - Verify (AB)-1 = B-1 A-1, A = [3 7 2 5], B =

Ex 4.5, 12 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.5, 12 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.5, 12 - Chapter 4 Class 12 Determinants - Part 4
Ex 4.5, 12 - Chapter 4 Class 12 Determinants - Part 5
Ex 4.5, 12 - Chapter 4 Class 12 Determinants - Part 6
Ex 4.5, 12 - Chapter 4 Class 12 Determinants - Part 7
Ex 4.5, 12 - Chapter 4 Class 12 Determinants - Part 8

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Transcript

Ex 4.5, 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)]

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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.