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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
    \[\left| \begin{matrix}    1+{{\sin }^{2}}\theta  & {{\sin }^{2}}\theta  & {{\sin }^{2}}\theta   \\    {{\cos }^{2}}\theta  & 1+{{\cos }^{2}}\theta  & {{\cos }^{2}}\theta   \\    4\sin 4\theta  & 4\sin 4\theta  & 1+4\sin 4\theta   \\ \end{matrix} \right|=0\] then \[\sin 4\theta \]equal to  [Orissa JEE 2005]

    A) 1/2

    B) 1

    C) -1/2

    D) -1

    Correct Answer: C

    Solution :

    \[\left| \,\begin{matrix}    1+{{\sin }^{2}}\theta  & {{\sin }^{2}}\theta  & {{\sin }^{2}}\theta   \\    {{\cos }^{2}}\theta  & 1+{{\cos }^{2}}\theta  & {{\cos }^{2}}\theta   \\    4\sin 4\theta  & 4\sin 4\theta  & 1+4\sin 4\theta   \\ \end{matrix}\, \right|=0\]Using \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\] Þ \[\left| \,\begin{matrix}    1 & 0 & {{\sin }^{2}}\theta   \\    -1 & 1 & {{\cos }^{2}}\theta   \\    0 & -1 & 1+4\sin 4\theta   \\ \end{matrix}\, \right|=0\] Þ \[2\,(1+2\sin 4\theta )=0\Rightarrow \sin 4\theta =\frac{-1}{2}\].


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