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JEE Main & Advanced Mathematics Determinants & Matrices Question Bank Expansion of determinants, Solution of equation in the form of determinants and properties of determinants

  • question_answer
    \[\left| \,\begin{matrix}    1 & 1 & 1  \\    \cos (nx) & \cos (n+1)x & \cos (n+2)x  \\    \sin (nx) & \sin (n+1)x & \sin (n+2)x  \\ \end{matrix}\, \right|\] is not depend  [RPET 2000]

    A) On x

    B) On n

    C) Both on x and n

    D) None of these

    Correct Answer: B

    Solution :

    \[\Delta =\left| \,\begin{matrix}    1 & 1 & 1  \\    \cos nx & \cos (n+1)x & \cos (n+2)x  \\    \sin nx & \sin (n+1)x & \sin (n+2)x  \\ \end{matrix}\, \right|\] Applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{3}}-(2\cos x){{C}_{2}}\] \[\Delta =\left| \,\begin{matrix}    2(1-\cos x) & 1 & 1  \\    0 & \cos (n+1)x & \cos (n+2)x  \\    0 & \sin (n+1)x & \sin (n+2)x  \\ \end{matrix}\, \right|\] \[\Delta =2(1-\cos x)[\cos (n+1)x\sin (n+2)x\] \[-\cos (n+2)x\sin (n+1)x]\] \[\Delta =2(1-\cos x)\,[\sin (n+2-n-1)x]\]\[=2\sin x(1-\cos x)\] i.e., \[\Delta \] is independent of n.


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