Slide4.JPG

Slide5.JPG
Slide6.JPG

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


Transcript

Ex 4.4, 2 Find adjoint of each of the matrices. [■8(1&−1&2@2&3&5@−2&0&1)] Let A = [■8(1&−1&2@2&3&5@−2&0&1)] adj A = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] Step 1: Cacluating minors M11 = |■8(3&5@0&1)|= 3(1) – 0(5) = 3 – 0 = 3 M12 = |■8(2&5@−2&1)| = 2 – ( –10) = 2 + 10 = 12 M13 = |■8(2&3@−2&0)| = 0 – ( – 6) = 0 + 6 = 6 M21 = |■8(−1&2@0&1)| = − 1 – (0) = – 1 + 0 = – 1 M22 = |■8(1&2@−2&1)| = 1 – (-2)2 = 1 + 4 = 5 M23 = |■8(1&−1@−2&0)| = 0 – ( – 2) ( – 1) = 0 – (2) = – 2 M31 = |■8(−1&2@3&5)| = 5 ( – 1) – 6 = – 5 – 6 = – 11 M32 = |■8(1&2@2&5)| = 5 – 4 = 1 M33 = |■8(1&−1@2&3)| = 3 – ( – 2) = 3 + 2 = 5 Step 2: Calculating adjoint A11 = 〖"( –1)" 〗^(1+1) . M11 = ( –1)2 . 3 = 3 A12 = 〖"( –1)" 〗^"1+2" . M12 = 〖"( –1)" 〗^"3" . (12) = – 1 (12) = – 12 A13 = 〖( −1)〗^(1+3) . M13 = 〖( −1)〗^4 . ( 6) = 1(6) = 6 A21 = 〖( −1)〗^(2+1) . M21 = 〖( −1)〗^3 . (-1) = -1(-1) = 1 A22 = 〖( −1)〗^(2+2) . M22 = 〖( −1)〗^4 . (5)= 1(5) = 5 A23 = 〖(−1)〗^(2+3). M23 = 〖(−1)〗^5. ( –2) = –1 (–2) = 2 A31 = 〖(−1)〗^(3+1). M31 = 〖(−1)〗^4 . ( –11) = 1 (– 11) = – 11 A32 = 〖(−1)〗^(3+2) . M32 = 〖(−1)〗^5. (1) = ( –1) (1) = –1 A33 = 〖(−1)〗^(3+3) . M33 = ( –1)6 . ( 5) = 1 (5) = 5 Step 3: Calculating adjoint Hence, adj A = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] = [■8(3&1&−11@−12&5&−1@6&2&5)]

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.