Question

The system of equation $\alpha x+y+z=\alpha -1,x+\alpha y+z=\alpha -1,x+y+\alpha z=\alpha -1$ has no solution if $\alpha$ is

• A. either $-2$ or $1$
• B. $-2$
• C. $1$
• D. $2$

Answer 1

The correct option is B $-2$

Here, $D=\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|$
$\Rightarrow D=(\alpha -1{)}^2(\alpha +2)$
Taking $D=0$
$\Rightarrow (\alpha -1{)}^2(\alpha +2)=0$
$\Rightarrow \alpha =1,\alpha =-2$
At these values of $\alpha$, system can have either no solution or infinitely many solution.
For $\alpha =1$, equations takes the form $x+y+z=0$
Hence, infinitely many solution for $\alpha =1$.
For $\alpha =-2$,
$D_1=\left|\begin{array}{ccc}-3& 1& 1\\ -3& -2& 1\\ -3& 1& -2\end{array}\right|$
$D_1=27\ne 0$
So, $D=0$, at least one $D_1\ne 0$
Hence, system has no solution for $\alpha =-2$