Misc 3 - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Inverse of two matrices and verifying properties
Inverse of two matrices and verifying properties
Last updated at April 16, 2024 by Teachoo
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)]