Question

Let $A=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]$ and $x\ne O$ be a column matrix such that $AX=\lambda X$ then sum of square of all possible values of $\lambda$ is

• A. $0,0$
• B. $-1,2$
• C. $4,2$
• D. $6,3$

Answer 1

The correct option is B $-1,2$

$A=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]A-\lambda I=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right]-\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right]$

$=\left[\begin{array}{ccc}-\lambda & 1& 1\\ 1& -\lambda & 1\\ 1& 1& -\lambda \end{array}\right]$

$|A-\lambda I=|\Rightarrow \left[\begin{array}{ccc}-\lambda & 1& 1\\ 1& -\lambda & 1\\ 1& 1& -\lambda \end{array}\right]=0$

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

Or, $-{\lambda }^3+\lambda +\lambda +1+1+\lambda =0$

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

$\lambda =-1$

$\therefore (\lambda +1)$ is a facator $\Rightarrow {\lambda }^2(\lambda +1)-\lambda (\lambda +1)-2(\lambda +1)$

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

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

$\therefore \lambda =-1,2$