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JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    If the system of linear equations                 \[{{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}=6\]                 \[{{x}_{1}}+3{{x}_{2}}+5{{x}_{3}}=9\]                 \[2{{x}_{1}}+5{{x}_{2}}+a{{x}_{3}}=b\]                 is consistent and has infinite number of solutions, then:     JEE Main  Online Paper (Held On 22 April 2013)

    A)  \[a=8,b\] can be any real number

    B)  \[b=15,\]a cab be any real number

    C)  \[a=R-\{8\}\] and \[\operatorname{b}\in \operatorname{R}-[15]\]

    D)  \[\operatorname{a}=8,b=15\]

    Correct Answer: D

    Solution :

     Given system of equations can be written in matrix form as AX = B where              \[A=\left( \begin{matrix}    1 & 2 & 3  \\    1 & 3 & 5  \\    2 & 5 & a  \\ \end{matrix} \right)\]and \[B=\left( \begin{matrix}    6  \\    9  \\    b  \\ \end{matrix} \right)\] Since, system is consistent and has infinitely many solutions \[\therefore \] (adj. A) B = 0 \[\Rightarrow \]               \[\left( \begin{matrix}    3a-25 & 15-2a & 1  \\    10-a & a-6 & -2  \\    -1 & -1 & 1  \\ \end{matrix} \right)\left( \begin{matrix}    6  \\    9  \\    b  \\ \end{matrix} \right)=\left( \begin{matrix}    0  \\    0  \\    0  \\ \end{matrix} \right)\] \[\Rightarrow \] \[-6-9+b=0\Rightarrow b=15\] and \[6(10-a)+9(a-6)-2(b)=0\] \[\Rightarrow \]               \[60-60a+9a-54-30=0\] \[\Rightarrow \]    \[3a=24\Rightarrow a=8\] Hence, \[a=8,b=15.\]


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