Ex 4.5, 2

Transcript

Ex 4.5, 2 Find adjoint of each of the matrices. 1﷮−1﷮2﷮2﷮3﷮5﷮−2﷮0﷮1﷯﷯ Let A = 1﷮−1﷮2﷮2﷮3﷮5﷮−2﷮0﷮1﷯﷯ adj A = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A13﷮A23﷮A33﷯﷯ Step 1: Cacluating minors M11 = 3﷮5﷮0﷮1﷯﷯= 3(1) – 0(5) = 3 – 0 = 3 M12 = 2﷮5﷮−2﷮1﷯﷯ = 2 – ( –10) = 2 + 10 = 12 M13 = 2﷮3﷮−2﷮0﷯﷯ = 0 – ( – 6) = 0 + 6 = 6 M21 = −1﷮2﷮0﷮1﷯﷯ = − 1 – (0) = – 1 + 0 = – 1 M22 = 1﷮2﷮−2﷮1﷯﷯ = 1 – (-2)2 = 1 + 4 = 5 M23 = 1﷮−1﷮−2﷮0﷯﷯ = 0 – ( – 2) ( – 1) = 0 – (2) = – 2 M31 = −1﷮2﷮3﷮5﷯﷯ = 5 ( – 1) – 6 = – 5 – 6 = – 11 M32 = 1﷮2﷮2﷮5﷯﷯ = 5 – 4 = 1 M33 = 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 = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A13﷮A23﷮A33﷯﷯ = 3﷮1﷮−11﷮−12﷮5﷮−1﷮6﷮2﷮5﷯﷯