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

Answer 1

The correct option is C not -2

$\alpha x+y+z=\alpha -1$,
$x+\alpha y+z=\alpha -1$
$x+y+z\alpha =\alpha -1$

using determinants to solve system if equations

$\Delta =\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|$

solving the determinant

$=\alpha ({\alpha }^2-1)-1(\alpha -1)+1(1-\alpha )$

$=\alpha (\alpha -1)(\alpha +1)-1(\alpha -1)-1(\alpha -1)$

for no solutions, determinant should be equal to zero

$\Rightarrow (\alpha -1)[{\alpha }^2+\alpha -2]=(\alpha -1)[{\alpha }^2+2\alpha -\alpha -2]=0$

$(\alpha -1)\left[\alpha \right(\alpha +2)-1(\alpha +2\left)\right]=0$

so $\alpha =$-2 or 1

so answer is option C