Using properties of determinants, prove that   where α, β
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    Determinants

    Multiple Choice QuestionsShort Answer Type

    51.  Without expanding, prove that the following determinants vanish :
     
    open vertical bar table row cell straight a minus straight b end cell cell space space straight b minus straight c space end cell cell straight c minus straight a end cell row cell straight b minus straight c end cell cell space straight c minus straight a end cell cell space straight a minus straight b end cell row cell straight c minus straight a end cell cell space straight a minus straight b end cell cell space straight b minus straight c end cell end table close vertical bar
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    52. Without space expanding comma space prove space that space the space following space determinants space vanish space open vertical bar table row 1 cell space space space space space space bc end cell cell space space space space space straight a left parenthesis straight b plus straight c right parenthesis end cell row 1 cell space space space space space ca end cell cell space space space space space straight b left parenthesis straight c plus straight a right parenthesis end cell row 1 cell space space space space space space ab end cell cell space space space space space straight c left parenthesis straight a plus straight b right parenthesis end cell end table close vertical bar
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    53.

    Without expansion, prove that the following determinants vanish 
    open vertical bar table row cell begin inline style 1 over straight a end style end cell cell space space space space straight a squared end cell cell space space bc end cell row cell begin inline style 1 over straight b end style end cell cell space space space straight b squared end cell cell space space ca end cell row cell begin inline style 1 over straight c end style end cell cell space space straight c squared end cell cell space space ab end cell end table close vertical bar

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    54.  Without expansion, prove that the following determinants vanish
     
    open vertical bar table row cell begin inline style 1 over straight a end style end cell cell space straight a space end cell bc row cell begin inline style 1 over straight b end style end cell cell space straight b end cell cell space ca end cell row cell begin inline style fraction numerator 1 over denominator space straight c end fraction end style end cell cell space straight c end cell cell space ab end cell end table close vertical bar
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    55. Without expanding, prove that 
    open vertical bar table row 1 cell space straight omega end cell cell space space straight omega squared end cell row straight omega cell space space straight omega squared end cell 1 row cell straight omega squared end cell cell space 1 end cell cell space straight omega end cell end table close vertical bar

    vanishes, where ω, ω2 are imaginary cube roots of unity.
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    56. Without expansion, show that

    open vertical bar table row 0 cell space space space straight a end cell cell space space minus straight b end cell row cell negative straight a end cell cell space space 0 end cell cell space space straight c end cell row straight b cell space space straight c end cell cell space space 0 end cell end table close vertical bar space equals space 0
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    57. If A + B + C = π. Find value of

    open vertical bar table row cell sin space left parenthesis straight A plus straight B plus straight C right parenthesis end cell cell space sin space straight B end cell cell space space cos space straight C end cell row cell negative sin space straight B end cell 0 cell space tan space straight A end cell row cell cos left parenthesis straight A plus straight B right parenthesis end cell cell space minus tan space straight A end cell cell space 0 end cell end table close vertical bar
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    58. open vertical bar table row cell 2 straight y plus 4 end cell cell space space space space 5 straight y plus 7 end cell cell space 8 straight y plus straight a end cell row cell 3 straight y plus 5 end cell cell space space space 6 straight y plus 8 end cell cell space space 9 straight y plus straight b end cell row cell 4 straight y plus 6 end cell cell space space space 7 straight y plus 9 end cell cell space space 10 straight y plus straight c end cell end table close vertical barIf a, b, c are in A.P. find value of
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    59.  Using properties of determinants, prove that
     
     space open square brackets table row cell straight x minus 3 end cell cell space space straight x minus 4 space end cell cell space space straight x minus straight alpha end cell row cell straight x minus 2 end cell cell space space straight x minus 3 end cell cell space space straight x minus straight beta end cell row cell straight x minus 1 end cell cell space space straight x minus 2 end cell cell space space straight x minus straight gamma end cell end table close square brackets space equals 0
     
    where α, β, γ, are in A.P.

    space Let space increment space equals space open square brackets table row cell straight x minus 3 end cell cell space space straight x minus 4 space end cell cell space space straight x minus straight alpha end cell row cell straight x minus 2 end cell cell space space straight x minus 3 end cell cell space space straight x minus straight beta end cell row cell straight x minus 1 end cell cell space space straight x minus 2 end cell cell space space straight x minus straight gamma end cell end table close square brackets space space space space space space space space space space space space space equals space open square brackets table row cell straight x minus 3 end cell cell space space straight x minus 4 space end cell cell space space straight x minus straight alpha end cell row 0 cell space space 0 end cell cell space space minus 2 straight beta plus straight a plus straight gamma end cell row cell straight x minus 1 end cell cell space space straight x minus 2 end cell cell space space straight x minus straight gamma end cell end table close square brackets space comma space by space straight R subscript 2 space rightwards arrow space 2 space space straight R subscript 2 space minus straight R subscript 1 space minus straight R subscript 1 space space space space space space space space space space equals space space open square brackets table row cell straight x minus 3 end cell cell space space straight x minus 4 space end cell cell space space straight x minus straight alpha end cell row 0 0 0 row cell straight x minus 1 end cell cell space space straight x minus 2 end cell cell space space straight x minus straight gamma end cell end table close square brackets space space space space space space space space open square brackets table row cell therefore space 2 beta equals alpha plus gamma space as space straight alpha comma straight beta comma straight gamma space are space in space straight A. straight P space end cell end table close square brackets space space space space space space space space space space equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets table row cell therefore space each space element space of space second space row space is space zero end cell end table close square brackets
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    60. open vertical bar table row 1 cell space space space straight a space end cell cell space bc end cell row 1 cell space space straight b end cell cell space space ca end cell row 1 cell space space straight c end cell cell space space ab end cell end table close vertical bar space equals space open vertical bar table row 1 cell space space space straight a space end cell cell space space space straight a squared end cell row 1 cell space space straight b end cell cell space space straight b squared end cell row 1 cell space space straight c end cell cell space space straight c squared end cell end table close vertical bar spaceWithout expanding, prove that
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