J & K CET Engineering J and K - CET Engineering Solved Paper-2006

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

    A)  \[{{a}^{3}}+{{b}^{3}}\]

    B)  \[-({{a}^{3}}+{{b}^{3}})\]

    C)  \[2\,({{a}^{3}}+{{b}^{3}})\]

    D)  \[-2\,({{a}^{3}}+{{b}^{3}})\]

    Correct Answer: D

    Solution :

    Let \[A=\left| \begin{matrix}    a & b & a+b  \\    b & a+b & a  \\    a+b & a & b  \\ \end{matrix} \right|\] Applying \[({{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}})\] \[=2(a+b)\left| \begin{matrix}    1 & 1 & 1  \\    b & a+b & a  \\    a+b & a & b  \\ \end{matrix} \right|\] Applying \[({{C}_{1}}\to {{C}_{1}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}})\] \[=2(a+b)\left| \begin{matrix}    0 & 0 & 1  \\    -a & b & a  \\    b & a-b & b  \\ \end{matrix} \right|\] \[=2(a+b)[1\{-a(a-b)-{{b}^{2}}\}]\] \[=2(a+b)(-{{a}^{2}}+ab-{{b}^{2}})\] \[=-2({{a}^{3}}+{{b}^{3}})\]


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