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 $\alpha =$

• A. -2
• B. either -2 or 1
• C. not -2
• D. 1

Answer 1

The correct option is A -2

$\alpha x+y+z=\alpha -1$
$x+\alpha y+z=\alpha -1$
$x+y+\alpha z=\alpha -1$
$\Delta =\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|$
$=\alpha ({\alpha }^2-1)-1(\alpha -1)+1(1-\alpha )$
$=\alpha (\alpha -1)(\alpha +1)-1(\alpha -1)-1(\alpha -1)$
System of equations has no solutions
$\Rightarrow (\alpha -1)({\alpha }^2+\alpha -1-1)=0$
$\Rightarrow (\alpha -1)({\alpha }^2+\alpha -2)=0$
$\Rightarrow (\alpha -1)({\alpha }^2+2\alpha -\alpha -2)=0$
$\Rightarrow (\alpha -1{)}^2(\alpha +2)=0$
$\Rightarrow \alpha =-2,1$
But $\alpha \ne 1$, as otherwise system is homogeneous and has infinite solutions, so $\alpha =-2$.