• # 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)  $\operatorname{a}=8,b=15$

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, $a=8,b=15.$