Transpose of a matrix
Last updated at April 16, 2024 by Teachoo
Ex 3.3, 6 If (i) A = [■8(cos𝛼&sin𝛼@−sin𝛼&cos𝛼 )] , then verify that A’A = I Solving L.H.S. A’A Given A = [■8(cos𝛼&sin𝛼@−sin𝛼&cos𝛼 )] So, A’ = [■8(𝐜𝐨𝐬𝜶&−𝐬𝐢𝐧𝜶@𝐬𝐢𝐧𝜶&𝐜𝐨𝐬𝜶 )] A’ A = [■8(cos𝛼&〖−sin〗𝛼@sin𝛼&cos𝛼 )] [■8(cos𝛼&sin𝛼@−sin𝛼&cos𝛼 )] = [■8(cos𝛼.cos𝛼+〖(−sin〗〖𝛼)〖(−sin〗〖𝛼)〗 〗&cos𝛼 〖.sin〗𝛼+〖(−sin〗〖𝛼)cos𝛼 〗@sin𝛼. cos𝛼+cos〖𝛼 〖(−sin〗〖𝛼)〗 〗&sin𝛼.sin𝛼+cos〖𝛼 .cos𝛼 〗 )] = [■8(cos2𝛼+sin2𝛼&sin〖𝛼 cos〖𝛼−sin〖𝛼 cos𝛼 〗 〗 〗@sin𝛼 cos〖𝛼−sin𝛼 〗 cos𝛼&sin2𝛼+cos2 a)] = [■8(𝐜𝐨𝐬𝟐𝜶+𝐬𝐢𝐧𝟐 𝜶&𝟎@𝟎&𝐬𝐢𝐧𝟐𝜶+𝐜𝐨𝐬𝟐 𝒂)] Using sin2 θ + cos2 θ = 1 = [■8(1&0@0&1)] = I = R.H.S Hence L.H.S = R.H.S Hence Proved