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JEE Main & Advanced AIEEE Solved Paper-2008

  • question_answer
    Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that \[x=cy+bz=az+cx\] and \[z=bx+ay\]. Then \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc\] is equal to       AIEEE  Solved  Paper-2008

    A) 0                             

    B) 1                             

    C) 2                             

    D) - 1

    Correct Answer: B

    Solution :

    According to given condition\[\left| \begin{matrix}    1 & -c & -b  \\    c & -1 & a  \\    b & a & -1  \\ \end{matrix} \right|=0\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc=1\]


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