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 solutions, if $\alpha$ is

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

Answer 1

Answer: (B)

$-2$

For no soIution $\left|A\right|=0$ and $\left(adjA\right)\left(B\right)\ne 0$

Now $\left|A\right|=0\Rightarrow \left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|=0$

$\Rightarrow {\alpha }^3-3\alpha +2=0\Rightarrow (\alpha -1{)}^2(\alpha +2)=0\Rightarrow \alpha =1,-2$

But for $\alpha =1,\left|A\right|=0$ and $\left(adjA\right)\left(B\right)=0\Rightarrow \text{for}\alpha =1$ there exist infintiely many solution.

For $\alpha =-2,$ $\left|A\right|=0$ but $\left(adjA\right)\left(B\right)\ne 0\Rightarrow \exists$ no solution.
Hence, option B.