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For a given matrix A = open square brackets table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close square brackets which of the following statement holds good:

  1. A = A–1 straight for all theta element ofR    
  2. A is symmetric, for θ = (2n +1)fraction numerator pi over denominator 2 end fraction, n element ofI    
  3. A is orthogonal matrix for θelement of R    
  4. A is skew symmetric, for θ = npi, n element of I    

The correct answer is: A is orthogonal matrix for θelement of R


    For θ = (2n +1)fraction numerator pi over denominator 2 end fraction
    A=open square brackets table row 0 cell negative-or-plus 1 end cell row cell plus-or-minus 1 end cell 0 end table close square bracketswhich is not symmetric for θ = npi
    A = open square brackets table row cell plus-or-minus 1 end cell 0 row 0 cell plus-or-minus 1 end cell end table close square bracketswhich is not skew symmetric
    A2 = open square brackets table row cell cos invisible function application 2 theta end cell cell negative sin invisible function application 2 theta end cell row cell sin invisible function application 2 theta end cell cell cos invisible function application 2 theta end cell end table close square brackets≠ I so A ≠ A–1

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