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Ex 3.3, 6 - Verify that A’A = I if A = [cos a sin a - Transpose of a matrix

  1. Chapter 3 Class 12 Matrices
  2. Serial order wise
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Ex 3.3, 6 If (i) A = [■8(cos⁡𝛼&sin⁡𝛼@−sin⁡𝛼&cos⁡𝛼 )] , then verify that A’A = I Taking L.H.S. A’A Given A = [■8(cos⁡𝛼&sin⁡𝛼@−sin⁡𝛼&cos⁡𝛼 )] So, A’ = [■8(cos⁡𝛼&−sin⁡𝛼@sin⁡𝛼&cos⁡𝛼 )] A’ A = [■8(cos⁡𝛼&〖−sin〗⁡𝛼@sin⁡𝛼&cos⁡𝛼 )] [■8(cos⁡𝛼&sin⁡𝛼@−sin⁡𝛼&cos⁡𝛼 )] = [■8(cos⁡𝛼.cos⁡𝛼+〖(−sin〗⁡〖𝛼)〖(−sin〗⁡〖𝛼)〗 〗&cos⁡𝛼 〖.sin〗⁡𝛼+〖(−sin〗⁡〖𝛼)cos⁡𝛼 〗@sin⁡𝛼. cos⁡𝛼+cos⁡〖𝛼 〖(−sin〗⁡〖𝛼)〗 〗&sin⁡𝛼.sin⁡𝛼+cos⁡〖𝛼 .cos⁡𝛼 〗 )] = [■8(cos2⁡𝛼+sin2𝛼&sin⁡〖𝛼 cos⁡〖𝛼−sin⁡〖𝛼 cos⁡𝛼 〗 〗 〗@sin⁡𝛼 cos⁡〖𝛼−sin⁡𝛼 〗 cos𝛼&sin2⁡𝛼+cos2 a)] = [■8(cos2⁡𝛼+sin2 𝛼&0@0&sin2⁡𝛼+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 Ex 3.3, 6 (ii) If A = [■8(sin⁡𝛼&cos⁡𝛼@−cos⁡𝛼&sin⁡𝛼 )] , then verify that A’ A = I Taking L.H.S A’ A Given A = [■8(sin⁡𝛼&cos⁡𝛼@−cos⁡𝛼&sin⁡𝛼 )] So, A’ = [■8(sin⁡𝛼&〖−cos〗⁡𝛼@cos⁡𝛼&sin⁡𝛼 )] A’ A = [■8(sin⁡𝛼&〖−cos〗⁡𝛼@cos⁡𝛼&sin⁡𝛼 )] [■8(sin⁡𝛼&cos⁡𝛼@〖−cos〗⁡𝛼&sin⁡𝛼 )] = [■8(sin⁡𝛼 〖.sin〗⁡𝛼+〖(−cos〗⁡〖𝛼)〖(−cos〗⁡〖𝛼)〗 〗&sin⁡𝛼 〖.cos〗⁡𝛼+〖(−cos〗⁡〖𝛼)〖(sin〗⁡〖𝛼)〗 〗@cos⁡𝛼 〖.sin〗⁡𝛼+sin⁡〖𝛼 〖(−cos〗⁡〖𝛼)〗 〗&cos⁡𝛼 〖.cos〗⁡𝛼+sin⁡〖𝛼 〖.sin〗⁡𝛼 〗 )] = [■8(sin2⁡𝛼+cos2𝛼&sin⁡〖𝛼 cos⁡〖𝛼−cos⁡〖𝛼 sin⁡𝛼 〗 〗 〗@cos⁡𝛼 sin⁡〖𝛼−sin⁡𝛼 〗 cos𝛼&cos2⁡𝛼+sin2𝛼)] = [■8(sin2⁡𝛼+cos2𝛼&0@0&cos2⁡𝛼+sin2𝛼)] Using sin2 θ + cos2 θ = 1 = [■8(1&0@0&1)] = I = R.H.S Hence L.H.S = R.H.S Hence Proved

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