Question

The system of equations $\begin{array}{c}\alpha x+y+z=\alpha -1\\ x+\alpha y+z=\alpha -1\\ x+y+\alpha z=\alpha -1\end{array}$ has no solution, is $\alpha$ is

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

Answer 1

The correct option is D -2

Given system of equations $AX=B$
where $A=\left[\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}\alpha -1\\ \alpha -1\\ \alpha -1\end{array}\right]$
For no solution,
$D=0$
$\Rightarrow D=\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|=0$
$\Rightarrow (\alpha -1)({\alpha }^2+\alpha -2)=0$
$\Rightarrow (\alpha -1{)}^2(\alpha +2)=0$
$\Rightarrow \alpha =1,-2$
Now, for $\alpha =1$
$D_1=\left|\begin{array}{ccc}0& 1& 1\\ 0& 1& 1\\ 0& 1& 0\end{array}\right|$
$\Rightarrow D_1=0$
Also, $D_2=\left|\begin{array}{ccc}1& 0& 1\\ 1& 0& 1\\ 1& 0& 0\end{array}\right|=0$
$D_3=\left|\begin{array}{ccc}1& 1& 0\\ 1& 1& 0\\ 1& 1& 0\end{array}\right|=0$
So, $D=D_1=D_2=D_3=0$
Hence, at $\alpha =1$ , system has infinitely many solution.
Now, for $\alpha =-2$
$D_1=\left|\begin{array}{ccc}-3& 1& 1\\ -3& -2& 1\\ -3& 1& -2\end{array}\right|=-27\ne 0$
Hence, at least one $D_1,D_2,D_3$ is non-zero.