Question

The system of equations,
$ax-y-z=\alpha -1,x-ay-z=\alpha -1x-y-az=\alpha -1$
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.
$\left|\begin{array}{ccc}\alpha & -1& -1\\ 1& -\alpha & -1\\ 1& -1& -\alpha \end{array}\right|$
$or\alpha ({\alpha }^2-1)-1(\alpha -1)+1(1-\alpha )=0$
$or\alpha ({\alpha }^2-1)-2\alpha +2=0$
$or\alpha (\alpha -1)(\alpha +1)-2(\alpha -1)=0$
$or(\alpha -1)({\alpha }^2+\alpha -2)=0$
$or(\alpha -1)(\alpha +2)(\alpha -1)=0$
$or(\alpha -1{)}^2(\alpha +2)=0$
$or\alpha =1,-2$

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 $\alpha =-2.$