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Ex 4.5, 2
Ex 4.5, 3 Important
Ex 4.5, 4 Important
Ex 4.5, 5
Ex 4.5, 6 Important
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Ex 4.5, 9
Ex 4.5, 10 Important
Ex 4.5, 11 Important
Ex 4.5, 12 You are here
Ex 4.5, 13
Ex 4.5, 14 Important
Ex 4.5, 15 Important
Ex 4.5, 16
Ex 4.5, 17 (MCQ) Important
Ex 4.5, 18 (MCQ) Important
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)]