Question

If the system of equations $\lambda x_1+x_2+x_3=1,x_1+\lambda x_2+x_3=1,x_1+x_2+\lambda x_3=1$ is consistance, then $\lambda$ can be

• A. $5$
• B. $-2/3$
• C. $-3$
• D. None of these

Answer 1

The correct option is D None of these

Using Cramer's Rule,

To be consistent eqn,

$△={△}_1={△}_2={△}_3=0$

$△=\left|\begin{array}{ccc}\lambda & 1& 1\\ 1& \lambda & 1\\ 1& 1& \lambda \end{array}\right|=0$

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

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

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

$\lambda =1,-2---\left(1\right)$

${△}_1=\left|\begin{array}{ccc}1& 1& 1\\ 1& \lambda & 1\\ 1& 1& \lambda \end{array}\right|=0$

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

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

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

${△}_2=\left|\begin{array}{ccc}\lambda & 1& 1\\ 1& 1& 1\\ 1& 1& \lambda \end{array}\right|=0$

$\Rightarrow \lambda (\lambda -1)-1(\lambda -1)+1\left(0\right)=0$

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

$\Rightarrow \lambda =1---\left(3\right)$

${△}_3=\left|\begin{array}{ccc}\lambda & 1& 1\\ 1& \lambda & 1\\ 1& 1& 1\end{array}\right|=0$

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

From options,$\left(d\right)$ None of these Ans.