# Misc 7 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 7 If A-1 = 3−11−156−55−22 and B = 12−2−1300−21 , Find (AB)-1 We know that (AB)-1 = B-1 A-1 Calculating B-1 We know that B-1 = 1|B| adj (B) exists if |B| ≠ 0 |B| = 12−2−1300−21 = 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 = A11 A21 A31 A12 A22 A32 A13 A23 A33 B = 12−2−1300−21 M11 = 30−21 =3(1) – (-2)0 = 3 M12 = −1001 = -1(1) – 0(0)= –1 M13 = −130−2 =(-1)(-2) – 0(3)= 2 M21 = 2−2−21 =2(1)–(-2)(-2)= -2 M22 = 1−201 = 1(1) – 0(-2) = 1 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) = 326112225 Now, B-1 = 1|B| adj (B) Putting values = 11 326112225 = 326112225 Now, (AB)-1 = B-1 A-1 = 326112225 3−11−156−55−22 = 3 3+2 −15+6(−5)3 −1+2 6+6(−2)3 1+2 −5+6(2)1 3+1 −15+2(−5)1 −1+1 6+2(−2)1 1+1 −5+2(2)2 3+2 −15+5(−5)2 −1+2 6+5(−2)2 1+2 −5+5(2) = 9−30+30−3+12−123−10+123−15+10−1+6−41−5+46−30+25−2+12−102−10+10 = 𝟗−𝟑𝟓−𝟐𝟏𝟎𝟏𝟎𝟐

Chapter 4 Class 12 Determinants

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.