Question

If the system of equations
$x+y+z=22x+4y-z=63x+2y+\lambda z=\mu$
has infinitely many solutions, then

• A. $\lambda -2\mu =-5$
• B. $2\lambda +\mu =14$
• C. $\lambda +2\mu =14$
• D. 2$\lambda -\mu =5$

Answer 1

The correct option is B $2\lambda +\mu =14$

Since system of equations has infinitely many solutions, therefore
$D=0$
$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 2& 4& -1\\ 3& 2& \lambda \end{array}\right|=0$
$\Rightarrow 1(4\lambda +2)-1(2\lambda +3)+1(4-12)=0$
$\Rightarrow 4\lambda +2-2\lambda -3-8=0$
$\Rightarrow 2\lambda =9$
$\therefore \lambda =\frac{9}{2}$
and
$D_1=\left|\begin{array}{ccc}2& 1& 1\\ 6& 4& -1\\ \mu & 2& \frac{9}{2}\end{array}\right|=0$
$\Rightarrow 2(18+2)-1(27+\mu )+1(12-4\mu )=0$
$\Rightarrow 40-27-\mu +12-4\mu =0$
$\therefore \mu =5$
Now,
$\lambda -2\mu =-\frac{11}{2}$
$2\lambda +\mu =14$
$\lambda +2\mu =\frac{29}{2}$
2$\lambda -\mu =4$