Question

For the system of linear equations $\alpha x+y+z=1,x+\alpha y+z=1,x+y+\alpha z=\beta$, which one of the following statements is NOT correct ?

• A. It has infinitely many solutions if $\alpha =2$ and $\beta =-1$
• B. $x+y+z=\frac{3}{4}$ if $\alpha =2$ and $\beta =1$
• C. It has infinitely many solutions if $\alpha =1$ and $\beta =1$
• D. It has no solution if $\alpha =-2$ and $\beta =1$

Answer 1

Answer: (A)

$\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|=0$
$\alpha \left({\alpha }^2-1\right)-1(\alpha -1)+1(1-\alpha )=0$
${\alpha }^3-3\alpha +2=0$
${\alpha }^2(\alpha -1)+\alpha (\alpha -1)-2(\alpha -1)=0$
$(\alpha -1)\left({\alpha }^2+\alpha -2\right)=0$
$\alpha =1,\alpha =-2,1$
$\text{ For }\alpha =1,\beta =1$

${\Delta }_1=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right|=3-1-1\Rightarrow x=\frac{1}{4}$
${\Delta }_2=\left|\begin{array}{ccc}2& 1& 1\\ 1& 1& 1\\ 1& 1& 2\end{array}\right|=2-1=1\Rightarrow y=\frac{1}{4}$
${\Delta }_3=\left|\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 1\end{array}\right|=2-1=1\Rightarrow z=\frac{1}{4}$
For $\alpha =2\Rightarrow$ unique solution