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)

$\begin{vmatrix}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{vmatrix}=0$
$ \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=\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{vmatrix}=3-1-1 \Rightarrow x =\frac{1}{4}$
$\Delta_2=\begin{vmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{vmatrix}=2-1=1 \Rightarrow y =\frac{1}{4} $
$\Delta_3=\begin{vmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{vmatrix}=2-1=1 \Rightarrow z =\frac{1}{4} $
For $\alpha=2 \Rightarrow$ unique solution