Question

If the system of linear equations
$x+y+z=5x+2y+2z=6x+3y+\lambda z=\mu ,$
$(\lambda ,\mu \in R),$ has infinitey many solutions, then the value of $\lambda +\mu$ is :

• A. $12$
• B. $10$
• C. $9$
• D. $7$

Answer 1

The correct option is B $10$

$D=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 2\\ 1& 3& \lambda \end{array}\right|$

$C_2\to C_2-C_1,C_3\to C_3-C_1$

$=\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& 2& \lambda -1\end{array}\right|$
$=\lambda -1-2$

$D_1=\left|\begin{array}{ccc}5& 1& 1\\ 6& 2& 2\\ \mu & 3& \lambda \end{array}\right|=\left|\begin{array}{ccc}5& 1& 1\\ -4& 0& 0\\ \mu & 3& \lambda \end{array}\right|=4\lambda -12$

$D_2=\left|\begin{array}{ccc}1& 5& 1\\ 1& 6& 2\\ 1& \mu & \lambda \end{array}\right|=\lambda -\mu +4$

Similaly, $D_3=\mu -7$
For infinitely many solutions,
$D=D_1=D_2=D_3=0$
$\Rightarrow \lambda =3,\lambda -\mu =-4,\mu =7$
$\therefore \lambda +\mu =10$