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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 3 If A-1 = [■8(3&−1&1@−15&6&−5@5&−2&2)] and B = [■8(1&2&−2@−1&3&0@0&−2&1)] , Find (AB)-1 We know that (AB)−1 = B−1 A−1 We are given A-1 , so calculating B−1 Calculating B−1 We know that B−1 = 1/(|B|) adj (B) exists if |B| ≠ 0 |B| = |■8(1&2&−2@−1&3&0@0&−2&1)| = 1 (3 – 0) – 2(– 1 – 0) –2 (2 – 0) = 1 (3) –2 (–1) –2(2) = 3 + 2 – 4 = 1 Since |B| ≠ 0 Thus, B-1 exists Calculating adj B Now, adj B = [■8(A_11&A_21&A_31@A_12&A_22&A_32@A_13&A_23&A_33 )] B = [■8(1&2&−2@−1&3&0@0&−2&1)] M11 = |■8(3&0@−2&1)| =3(1) – (–2)0 = 3 M12 = |■8(−1&0@0&1)| = -1(1) – 0(0)= –1 M13 = |■8(−1&3@0&−2)| =(–1)(–2) – 0(3)= 2 M21 = |■8(2&−2@−2&1)| =2(1)–(–2)(–2)= –2 M22 = |■8(1&−2@0&1)| = 1(1) – 0(−2) = 1 M23 = |■8(1&2@0&−2)| = 1(-2) – 0(2) = –2 M31 = |■8(2&−2@3&0)| = 2(0) – 3(−2) = 6 M32 = |■8(1&−2@−1&0)| =1(0)–(–1)(–2)= –2 M33 = |■8(1&2@−1&3)| = 1(3) – (–1)2 = 5 Now, A11 = (–1)1+1 M11 = (–1)2 . 3 = 3 A12 = (–1)1+2 M12 = (–1)3 (–1) = 1 A13 = (–1)1+3 M13 = (–1)4 2 = 2 A21 = (–1)2+1 M21 = (–1)3 (–2) = 2 A22 = (–1)2+2 M22 = (–1)4 . 1 = 1 A23 = (–1)2+3 M23 = (–1)5 (–2) = 2 A31 = (–1)3+1 M31 = (–1)4 . 6 = 6 A32 = (–1)3+2 M32 = (–1)5 (–2) = 2 A33 = (–1)3+3 M33 = (–1)6 . 5 = 5 Thus, adj (B) = [■8(3&2&6@1&1&2@2&2&5)] Now, B−1 = 1/(|B|) adj (B) Putting values = 1/1 [■8(3&2&6@1&1&2@2&2&5)] = [■8(3&2&6@1&1&2@2&2&5)] Also, (AB)-1 = B−1 A−1 = [■8(3&2&6@1&1&2@2&2&5)] [■8(3&−1&1@−15&6&−5@5&−2&2)] = [■8(3(3)+2(⤶7−15)+6(−5)&3(−1)+2(6)+6(−2)&3(1)+2(−5)+6(2)@1(3)+1(⤶7−15)+2(−5)&1(−1)+1(6)+2(−2)&1(1)+1(−5)+2(2)@2(3)+2(⤶7−15)+5(−5)&2(−1)+2(6)+5(−2)&2(1)+2(−5)+5(2))] = [■8(9−30+30&−3+12−12&3−10+12@3−15+10&−1+6−4&1−5+4@6−30+25&−2+12−10&2−10+10)] = [■8(𝟗&−𝟑&𝟓@−𝟐&𝟏&𝟎@𝟏&𝟎&𝟐)]

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.