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Misc 7 - Find (AB)-1, A-1 = [3 -1 1 - Class 12 Determinants

Misc 7 - Chapter 4 Class 12 Determinants - Part 2
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Misc 7 If A-1 = [■8(3&−1&[email protected]−15&6&−[email protected]&−2&2)] and B = [■8(1&2&−[email protected]−1&3&[email protected]&−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&−[email protected]−1&3&[email protected]&−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&[email protected]_12&A_22&[email protected]_13&A_23&A_33 )] B = [■8(1&2&−[email protected]−1&3&[email protected]&−2&1)] M11 = |■8(3&[email protected]−2&1)| =3(1) – (–2)0 = 3 M12 = |■8(−1&[email protected]&1)| = -1(1) – 0(0)= –1 M13 = |■8(−1&[email protected]&−2)| =(–1)(–2) – 0(3)= 2 M21 = |■8(2&−[email protected]−2&1)| =2(1)–(–2)(–2)= –2 M22 = |■8(1&−[email protected]&1)| = 1(1) – 0(−2) = 1 M23 = |■8(1&[email protected]&−2)| = 1(-2) – 0(2) = –2 M31 = |■8(2&−[email protected]&0)| = 2(0) – 3(−2) = 6 M32 = |■8(1&−[email protected]−1&0)| =1(0)–(–1)(–2)= –2 M33 = |■8(1&[email protected]−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&[email protected]&1&[email protected]&2&5)] Now, B−1 = 1/(|B|) adj (B) Putting values = 1/1 [■8(3&2&[email protected]&1&[email protected]&2&5)] = [■8(3&2&[email protected]&1&[email protected]&2&5)] Also, (AB)-1 = B−1 A−1 = [■8(3&2&[email protected]&1&[email protected]&2&5)] [■8(3&−1&[email protected]−15&6&−[email protected]&−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−[email protected]−15+10&−1+6−4&1−[email protected]−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.