Discussion Forum

 The set of all values of   $\lambda$  for  which system of linear  equations ,$2X_{1}-2X_{2}+X_{3}=\lambda X_{1}$, $2X_{1}-3X_{2}+2X_{3}=\lambda X_{2}$   and  $-X_{1}+2X_{2}=\lambda X_{3}$   has   a non-trival solution.

Answer:

Explanation:

 Given system of linear equations

     $2X_{1}-2X_{2}+X_{3}=\lambda X_{1},$

$\Rightarrow$             $(2-\lambda)X_{1}-2X_{2}+X_{3}=0        .......(i)$

     $2X_{1}-3X_{2}+2X_{3}=\lambda X_{2}$

  $\Rightarrow$     $2X_{1}-(3+\lambda)X_{2}+2X_{3}=0        .......(ii)$

                  $-X_{1}+2X_{2}=\lambda X_{3}$

   $\Rightarrow$  $-X_{1}+2X_{2}-\lambda X_{3}=0$                         .......(iii)

 Since, the system has non-trival solution.

      $\therefore$           $\begin{bmatrix}2-\lambda & -2&1 \\2 & -(3+\lambda)& 2  \\ -1&2&-\lambda \end{bmatrix}=0$

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

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

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

$\Rightarrow  (\lambda-1)[(2-\lambda)(\lambda+4)-5]=0$

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

 $\Rightarrow  (\lambda-1)[(\lambda-1)(\lambda+3)]=0$

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

  $\Rightarrow  \lambda=1,1,-3