Check sibling questions

Example 25 - Verify (AB)-1 = B-1 A-1, if A = [2 3 1 -4] - Inverse of two matrices and verifying properties

Example 25 - Chapter 4 Class 12 Determinants - Part 2
Example 25 - Chapter 4 Class 12 Determinants - Part 3
Example 25 - Chapter 4 Class 12 Determinants - Part 4
Example 25 - Chapter 4 Class 12 Determinants - Part 5
Example 25 - Chapter 4 Class 12 Determinants - Part 6
Example 25 - Chapter 4 Class 12 Determinants - Part 7

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Transcript

Example 25 If A = 2 3 1 4 and B = 1 2 1 3 , then verify that (AB)-1 = B-1 A-1 Taking L .H.S (AB) 1 First calculating AB AB = 2 3 1 4 1 2 1 3 = 2 1 +3( 1) 2 2 +3(3) 1 1 +( 4)( 1) 1 2 + 4 3 = 2 3 4+9 1+4 2 12 = 1 5 5 14 Now, (AB)-1 = 1 |AB| adj (AB) exists if |AB| 0 |AB| = 1 5 5 14 = (-1)(-14) 5(5) = 14 25 = 11 Since |AB| 0 (AB)-1 exists AB = 1 5 5 14 adj (AB) = 1 5 5 14 = 14 5 5 1 Now, (AB) 1 = 1 |AB| adj (AB) Putting values = 1 11 14 5 5 1 = 1 11 14 5 5 1 Taking R.H.S B-1A-1 First Calculating B 1 B = 1 2 1 3 B = 1 |B| adj (B) exists if |B| 0 |B| = 1 2 1 3 = 3 2 = 1 Since |B| 0 , B-1 exist B = 1 2 1 3 adj (B) = 1 2 1 3 = 3 2 1 1 Now, (B) 1 = 1 |B| adj (B) Putting values = 1 1 3 2 1 1 = 3 2 1 1 Finding A-1 A-1 = 1 |A| adj (A) exists if |A| 0 |A| = 2 3 1 4 = 2 ( 4) 1( 3) = 8 3 = 11 Since |A| 0 , A 1 exists A = 2 3 1 4 adj (A) = 2 3 1 4 = 4 3 1 2 Now, A-1 = 1 |A| adj (A) = 1 11 4 3 1 2 = 1 11 4 3 1 2 Thus, B-1A-1 = 3 2 1 1 1 11 4 3 1 2 = 1 11 3 2 1 1 4 3 1 2 = 1 11 3 4 +2(1) 3 3 +2( 2) 1 4 +1(1) 1 3 +1( 2) = 1 11 12+2 9 4 4+1 3 2 = 1 11 14 5 5 1 = L.H.S L.H.S = R.H.S Hence proved

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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.