CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    If value of\[\left| \begin{matrix}    a & a+b & a+2b  \\    a+2b & a & a+b  \\    a+b & a+2b & a  \\ \end{matrix} \right|\]is equal to:

    A)  \[9{{a}^{2}}(a+b)\]                        

    B)  \[9{{b}^{2}}(a+b)\]

    C)  \[{{a}^{2}}(a+b)\]          

    D)         \[{{b}^{2}}(a+b)\]

    E)  \[9{{b}^{2}}(a-b)\]

    Correct Answer: B

    Solution :

    The determinant value of matrix \[3AB={{3}^{3}}(-1)(3)=-81\] \[\left| \begin{matrix}    a & a+b & a+2b  \\    a+2b & a & a+b  \\    a+b & a+2b & a  \\ \end{matrix} \right|\] \[=\left| \begin{matrix}    3(a+b) & 3(a+b) & 3(a+b)  \\    a+2b & a & a+b  \\    a+b & a+2b & a  \\ \end{matrix} \right|\]                                                 \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\] \[=3(a+b)\left| \begin{matrix}    1 & 0 & 0  \\    a+2b & -2b & -b  \\    a+2b & b & -b  \\ \end{matrix} \right|\]                                                 \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\]                                                 \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] \[=3(a+b)\left| \begin{matrix}    -2b & -b  \\    b & -b  \\ \end{matrix} \right|\] \[=3{{b}^{2}}(a+b)\left| \begin{matrix}    2 & 1  \\    -1 & 1  \\ \end{matrix} \right|\] \[=3{{b}^{2}}(a+b)(2+1)\] \[=9{{b}^{2}}(a+b)\]


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