Question

The values of $\lambda$ such that $(x,y,z)\ne (0,0,0)$ and $(\hat{i}+\hat{j}+3\hat{k})x+(3\hat{i}-3\hat{j}+\hat{k})y+(-4\hat{i}+5\hat{j})z=\lambda (x\hat{i}+y\hat{j}+z\hat{k})$ are

• A. $0,1$
• B. $0,-1$
• C. $1,-1$
• D. $0,1,-1$

Answer 1

Answer: (B)

$0,-1$

$\Rightarrow (x+3y-4z-\lambda x)\hat{i}+(x-3y+5z-\lambda y)\hat{j}+(3x+y-\lambda z)\hat{k}=0$

$\Rightarrow x+3y-4z-\lambda x=0$

$x-3y+5z-\lambda y=0$

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

$\left[\begin{array}{ccc}1-\lambda & 3& -4\\ 1& -3-\lambda & 5\\ 3& 1& -\lambda \end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

$\Rightarrow \left|\begin{array}{ccc}1-\lambda & 3& -4\\ 1& -3-\lambda & 5\\ 3& 1& -\lambda \end{array}\right|=0\left\{\begin{array}{c}\because \text{for non-trivial}\\ \text{soution of equation}\\ Ax=y\\ |A|=0\end{array}\right\}$

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

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

$=(1-\lambda )({\lambda }^2+3\lambda -5)-4+3\lambda +9-12\lambda =0$

$={\lambda }^2+3\lambda -5-{\lambda }^3-3{\lambda }^2+5\lambda -4+3\lambda +9-12\lambda =0$

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

$\Rightarrow {\lambda }^3+2{\lambda }^2+{\lambda }^2=0$

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

$\lambda =0,{\lambda }^2+2\lambda +1=0$

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

$\lambda =-1$

$\Rightarrow \lambda =0,-1$

$\therefore$ option $B$ is correct.