Question

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

• A. Either $-2$ or $1$
• B. $-2$
• C. $1$
• D. Not $-2$

Answer 1

The correct option is B $-2$

For no solution or infinitely many solutions, the condition to be satisfied is $\Delta =0$
$\Rightarrow \left|\begin{array}{ccc}\alpha & -1& -1\\ 1& -\alpha & -1\\ 1& -1& -\alpha \end{array}\right|=0$
$\Rightarrow \alpha ({\alpha }^2-1)-1(\alpha -1)+1(1-\alpha )=0$
$\Rightarrow \alpha ({\alpha }^2-1)-2\alpha +2=0$
$\Rightarrow \alpha (\alpha -1)(\alpha +1)-2(\alpha -1)=0$
$\Rightarrow (\alpha -1)({\alpha }^2+\alpha -2)=0$
$\Rightarrow (\alpha -1)(\alpha +2)(\alpha -1)=0$
$\Rightarrow (\alpha -1{)}^2(\alpha +2)=0$
$\Rightarrow \alpha =1,-2$

But for $\alpha =1,$ there are infinite solutions.
When $\alpha =-2,$ we have
$-2x-y-z=-3$
$x+2y-z=-3$
$x-y+2z=-3$
Adding all equations, we get $0=-9,$ which is not true.
Hence, there is no solution when $\alpha =-2.$