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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Integral power of iota, Algebraic operations and Equality of complex numbers

  • question_answer
    If \[\,\left| \begin{align}   & \,6i\,\,\,\,\,-3i\,\,\,\,\,\,\,\,\,1 \\  & \,\,4\,\,\,\,\,\,\,\,\,3i\,\,\,\,\,\,-1 \\  & \,20\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,i \\ \end{align} \right|\,\]=\[x+iy\], then (x, y) is [MP PET 2000]

    A) (3, 1)

    B) (1, 3)

    C) (0, 3)

    D) (0, 0)

    Correct Answer: D

    Solution :

    \[\left| \,\begin{matrix}    6i\,\,\, & -3i\,\,\, & \,\,1  \\    4\,\, & 3i & -1  \\    20\, & 3 & \,\,i  \\ \end{matrix}\, \right|\]=\[x+iy\] Þ  \[\left| \,\begin{matrix}    6i+4\,\,\,\, & 0\,\,\,\, & \,\,0  \\    4\,\,\, & 3i\,\,\,\, & -1  \\    20\,\,\, & 3\,\,\,\, & \,\,i  \\ \end{matrix}\, \right|=x+iy\] \[[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}]\] Þ  \[(6i+4)\,(3{{i}^{2}}+3)\]= \[x+iy\]  Þ  \[(6i+4)\,(-3+3)=x+iy\] Þ  \[x+iy=0\,\,=0+i.0\] Þ   \[(x,\,y)\,=(0,\,0)\].


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