

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 4.4
Ex 4.4, 2 You are here
Ex 4.4, 3 Important
Ex 4.4, 4 Important
Ex 4.4, 5
Ex 4.4, 6 Important
Ex 4.4, 7
Ex 4.4, 8
Ex 4.4, 9
Ex 4.4, 10 Important
Ex 4.4, 11 Important
Ex 4.4, 12
Ex 4.4, 13
Ex 4.4, 14 Important
Ex 4.4, 15 Important
Ex 4.4, 16
Ex 4.4, 17 (MCQ) Important
Ex 4.4, 18 (MCQ) Important
Last updated at June 11, 2023 by Teachoo
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)]