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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
    The values of the determinant \[\left| \,\begin{matrix}    1 & \cos (\alpha -\beta ) & \cos \alpha   \\    \cos (\alpha -\beta ) & 1 & \cos \beta   \\    \cos \alpha  & \cos \beta  & 1  \\ \end{matrix}\, \right|\] is [UPSEAT 2003]

    A) \[{{\alpha }^{2}}+{{\beta }^{2}}\]

    B) \[{{\alpha }^{2}}-{{\beta }^{2}}\]

    C) 1

    D) 0

    Correct Answer: D

    Solution :

    On solving the determinant, \[1\,(1-{{\cos }^{2}}\beta )-\cos (\alpha -\beta )\]\[[\cos (\alpha -\beta )-\cos \alpha \cos \beta ]\]\[+\cos \alpha [\cos \beta \cos (\alpha -\beta )-\cos \alpha ]\] \[=1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha -{{\cos }^{2}}(\alpha -\beta )\]\[+2\cos \alpha \cos \beta \cos (\alpha -\beta )\] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha +\cos (\alpha -\beta )\]\[(2\cos \alpha \cos \beta -\cos (\alpha -\beta ))\] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha +\cos (\alpha -\beta )\]\[[\cos (\alpha +\beta )+\cos (\alpha -\beta )-\cos (\alpha -\beta )]\] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha +\cos (\alpha -\beta )\cos (\alpha +\beta )\] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha {{\cos }^{2}}\beta -{{\sin }^{2}}\alpha {{\sin }^{2}}\beta \] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha (1-{{\cos }^{2}}\beta )-{{\sin }^{2}}\alpha {{\sin }^{2}}\beta \] = \[1-{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha {{\sin }^{2}}\beta -{{\sin }^{2}}\alpha {{\sin }^{2}}\beta \] = \[1-{{\cos }^{2}}\beta -{{\sin }^{2}}\beta \] = 0.


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