Question

Find all $\lambda \in R$such that $\left(x,y,z\right)\ne \left(0,0,0\right)$ and $\left(\hat{i}+\hat{j}+\hat{k}\right)x+\left(3\hat{i}-3\hat{j}+\hat{k}\right)y+\left(-4\hat{i}+5\hat{j}\right)z=\lambda \left(x\hat{i}+y\hat{j}+z\hat{k}\right)$

Answer 1

$\stackrel{\wedge }{i}(x+3y-4z)+\stackrel{\wedge }{j}(x-3y+5z)+\stackrel{\wedge }{k}(3x+y)=\stackrel{\wedge }{i}\left(\lambda x\right)+\stackrel{\wedge }{j}\left(\lambda y\right)+\stackrel{\wedge }{k}\left(\lambda z\right)$

$comparingbothsides$

$⟹x+3y-4z=\lambda x-\left(1\right)$

$⟹x-3y+5z=\lambda y-\left(2\right)$

$⟹3x+y=\lambda z-\left(3\right)$

$⟹Rearranging\left(1\right),\left(2\right)\&\left(3\right)$

$⟹x(1-\lambda )+y\left(3\right)+2(-4)=0-\left(4\right)$

$x\left(1\right)+y(-3-\lambda )+2\left(5\right)=0-\left(5\right)$

$x\left(3\right)+y\left(1\right)+2(-\lambda )=0-\left(6\right)$

$Weknowifa1x+a2y+a3z=0$

$b1x+b2y+b3z=0$

$c1x+c2y+c3z=0$

$⟹\left|\begin{array}{ccc}a1& a2& a3\\ b1& b2& b3\\ c1& c2& c3\end{array}\right|=0\therefore Here,\left|\begin{array}{ccc}1-\lambda & 3& -4\\ 1& -3-\lambda & 5\\ 3& 1& -\lambda \end{array}\right|=0$

$⟹(1-\lambda )\left[\right(3\lambda -{\lambda }^2)-5]-3(-\lambda -15)-4(1+9+3\lambda )=0$

$⟹(1-\lambda )({\lambda }^2-3\lambda -5)+3\lambda +45-40-12\lambda =0$

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

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

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

$⟹\lambda =0or{\lambda }^2+2\lambda +1=0$

$⟹{(\lambda +1)}^2=0$

$⟹\lambda =-1$

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