Ex 4.5, 12 - Verify (AB)-1 = B-1 A-1, A = [3 7 2 5], B = - Inverse of two matrices and verifying properties

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  1. Chapter 4 Class 12 Determinants
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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

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