Question

The value of $\lambda$ and $\mu$ for which the system of equations $x+y+z=6,x+2y+3z=10$ and $x+2y+\lambda z=\mu$ have no solution, are

• A. $\lambda =3,\mu \ne 10$
• B. $\lambda \ne 3,\mu =10$
• C. $\lambda \ne 3,\mu \ne 10$
• D. None of these

Answer 1

Answer: (A)

$\lambda =3,\mu \ne 10$

Given system of equation is
$x+y+z=6$
$x+2y+3z=10$
$x+2y+\lambda z=\mu$
If given system of equation has no solution, then $D=0$ and atleast one of the determinants $D_1,D_2,D_3\ne 0$
Here $D=0$
$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& \lambda \end{array}\right|=0$
$\Rightarrow 1(2\lambda -6)-1(\lambda -3)+1(2-2)=0$
$\Rightarrow 2\lambda -6-\lambda +3+0=0$
$\Rightarrow \lambda -3=0$
$\Rightarrow \lambda =3$
Also, $D_1=\left|\begin{array}{ccc}6& 1& 1\\ 10& 2& 3\\ \mu & 2& 2\end{array}\right|\ne 0$
$\Rightarrow 6(6-6)-(30-3\mu )+1(20-2\mu )\ne 0$
$\Rightarrow 0-30+3\mu +20-2\mu \ne 0$
$\Rightarrow \mu -10\ne 0$
$\Rightarrow \mu \ne 10$.