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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    If \[\left[ \begin{matrix}    1 & 2 & -1  \\    1 & x-2 & 1  \\    x & 1 & 1  \\ \end{matrix} \right]\]is singular, then the value of \[x\]is

    A)  2                                            

    B)  3

    C)  1                                            

    D)  0

    Correct Answer: A

    Solution :

    Let \[A=\left[ \begin{matrix}    1 & 2 & -1  \\    1 & x-2 & 1  \\    x & 1 & 1  \\ \end{matrix} \right]\] If the matrix A is singular, then \[|A|=0\]                                 \[\left| \begin{matrix}    1 & 2 & -1  \\    1 & x-2 & 1  \\    x & 1 & 1  \\ \end{matrix} \right|=0\]                 Expand with respect to \[{{R}_{1}}\]                 \[\Rightarrow \] \[1(x-2-1)-2(1-x)-1(1-{{x}^{2}}+2x)=0\] \[\Rightarrow \] \[x-3-2+2x-1+{{x}^{2}}-2x=0\] \[\Rightarrow \]               \[{{x}^{2}}+x-6=0\] \[\Rightarrow \]               \[(x-2)\,(x+3)=0\] \[\Rightarrow \]               \[x=2,\,-3\]


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