Question

The system of equations
$\lambda x+y+3z=0$
$2x+\mu y-z=0$
$5x+7y+z=0$
has infinitely many solutions in $R$. Then,

• A. $\lambda =2,\mu =3$
• B. $\lambda =1,\mu =2$
• C. $\lambda =1,\mu =3$
• D. $\lambda =3,\mu =1$

Answer 1

The correct option is C $\lambda =1,\mu =3$

$\left|\begin{array}{ccc}\lambda & 1& 3\\ 2& \mu & -1\\ 5& 7& 1\end{array}\right|=0\Rightarrow \lambda (\mu +7)-1(2+5)+3(14-5\mu )=0\Rightarrow \lambda \mu +7\lambda -7+42-15\mu =0$
$\Rightarrow \lambda \mu +7\lambda -15\mu +35=0$
$\Rightarrow \lambda (\mu +7)-15(\mu +7)=-140$
$\Rightarrow (\lambda -15)(\mu +7)=-140$

From options, only $\lambda =1,\mu =3$ satisfying above equation.