Ex 4.4, 12 - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Inverse of two matrices and verifying properties
Inverse of two matrices and verifying properties
Last updated at April 16, 2024 by Teachoo
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