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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
    If A, B, C be the angles of a triangle, then \[\left| \,\begin{matrix}    -1 & \cos C & \cos B  \\    \cos C & -1 & \cos A  \\    \cos B & \cos A & -1  \\ \end{matrix}\, \right|=\] [Karnataka CET 2002]

    A) 1

    B) 0

    C) \[\cos A\cos B\cos C\]

    D) \[\cos A+\cos B\cos C\]

    Correct Answer: B

    Solution :

    Given, Angles of a triangle = A, B and C. We know that as A + B + C = p, therefore \[A+B=\pi -C\] or \[\cos (A+B)=\cos (\pi -C)=-\cos C\] or \[\cos A\cos B-\sin A\sin B=-\cos C\]  \[\cos A\cos B+\cos C=\sin A\sin B\] and \[\sin (A+B)=\sin (\pi -C)=\sin C.\] Expanding the given determinant, we get \[\Delta =-(1-{{\cos }^{2}}A)+\cos C(\cos C+\cos A\cos B)\]                           \[+\cos B(\cos B+\cos A\cos C)\]          \[=-{{\sin }^{2}}A+\cos C(\sin A\sin B)+\cos B(\sin A\sin C)\]          \[=-{{\sin }^{2}}A+\sin A(\sin B\cos C+\cos B\sin C)\]      \[=-{{\sin }^{2}}A+\sin A\sin (B+C)\]\[=-{{\sin }^{2}}A+{{\sin }^{2}}A=0.\]


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