Question

The equations $x+3y-4z=$ $\lambda x$; $x-3y+5z=$ $\lambda y$; $3x+y+0=$ $\lambda z$ have infinite solutions then $\lambda$ =

• A. 0
• B. 1,-1
• C. 0,-1
• D. 0,1

Answer 1

Answer: (C)

0,-1

$(1-\lambda )x+3y-4z=0$

$x-(\lambda +2)y+5z$

$3x+y-\lambda z=0$

$\Rightarrow \left|\begin{array}{ccc}1-\lambda & 3& -4\\ 1& -(\lambda +2)& 5\\ 3& 1& -\lambda \end{array}\right|=0$

$\Rightarrow (1-\lambda )(\lambda (\lambda +2)-5)-3(-\lambda -15)-4(1+3(\lambda +3))=0$

$\Rightarrow (1-\lambda )({\lambda }^2+2\lambda -5)+3(\lambda +15)-4(3\lambda +10)=0$

$\Rightarrow {\lambda }^2+3\lambda -5-\lambda ({\lambda }^2+3\lambda -5)+3\lambda +45-12\lambda -40=0$

$\Rightarrow {\lambda }^2+3\lambda -5-{\lambda }^3-3{\lambda }^2+5\lambda +3\lambda +45-12\lambda -40=0$

$\Rightarrow -{\lambda }^3-1{\lambda }^2=0$

$\Rightarrow -{\lambda }^2(\lambda +1)=0$

$\Rightarrow \lambda =0$ or $-1$