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    Maths-
    General
    Easy

    Question

    Let A =open square brackets table row 1 cell sin invisible function application theta end cell 1 row cell – sin invisible function application theta end cell 1 cell sin invisible function application theta end cell row cell – 1 end cell cell – sin invisible function application theta end cell 1 end table close square brackets, where 0 ≤ θ < 2pi, then

    1. Det (A) = 0    
    2. Det A element of (0, straight infinity)    
    3. Det (A) element of [2, 4]    
    4. Det A element of [2, straight infinity)    

    hintHint:

    Using the determinant of the matrix find equation in terms of sin θ and find the range of sin squared space theta space and Find the range of determinant 

    The correct answer is: Det (A) element of [2, 4]

      Given,
      open square brackets table row 1 cell sin invisible function application theta end cell 1 row cell – sin invisible function application theta end cell 1 cell sin invisible function application theta end cell row cell – 1 end cell cell – sin invisible function application theta end cell 1 end table close square brackets space equals space 1 open vertical bar table row 1 cell sin invisible function application theta end cell row cell negative sin invisible function application theta end cell 1 end table close vertical bar space minus space sin invisible function application theta open vertical bar table row cell negative sin invisible function application theta end cell cell sin invisible function application theta end cell row cell negative 1 end cell 1 end table close vertical bar space plus 1 open vertical bar table row cell negative sin invisible function application theta end cell 1 row cell negative 1 end cell cell negative sin invisible function application theta end cell end table close vertical bar equals space 1 space plus space sin squared open parentheses theta close parentheses space minus space sin invisible function application theta left square bracket space 0 space right square bracket space plus space sin squared open parentheses theta close parentheses space plus 1 open vertical bar A close vertical bar space o r space D e t space left parenthesis A right parenthesis space equals space 2 space plus space 2 space sin squared open parentheses theta close parentheses space
      As range of sin squared space theta spacelies [0,1]  Then  Det (A) element of [2+0, 2 + 2(1) ]
      So, Det (A) element of [2, 4 ]

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