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For the system of linear equations $a x+y+z=1$, $x+a y+z=1, x+y+a z=\beta$, which one of the following statements is NOT correct ?
It has infinitely many solutions if $\alpha=2$ and $\beta=-1$
It has no solution if $\alpha=-2$ and $\beta=1$
$x+y+z=\frac{3}{4}$ if $\alpha=2$ and $\beta=1$
It has infinitely many solutions if $\alpha=1$ and $\beta=1$
Solution
$\left|\begin{array}{lll}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right|=0$
$\alpha\left(\alpha^2-1\right)-1(\alpha-1)+1(1-\alpha)=0$
$\alpha^3-3 \alpha+2=0$
$\alpha^2(\alpha-1)+\alpha(\alpha-1)-2(\alpha-1)=0$
$(\alpha-1)\left(\alpha^2+\alpha-2\right)=0$
$\alpha=1, \alpha=-2,1$
$\text { For } \alpha=1, \beta=1$
$\left.\begin{array}{l}x+y+z=1 \\ x+y+z=b\end{array}\right\} \text { infinite solution }$
For $\alpha=2, \beta=1$
$\begin{array}{l}\Delta=4 \\\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\1 & 2 & 1 \\1 & 1 & 2\end{array}\right|=3-1-1 \quad \Rightarrow x =\frac{1}{4} \\\Delta_2=\left|\begin{array}{lll}2 & 1 & 1 \\1 & 1 & 1 \\1 & 1 & 2\end{array}\right|=2-1=1\quad \Rightarrow y =\frac{1}{4} \\\Delta_3=\left|\begin{array}{ccc}2 & 1 & 1 \\1 & 2 & 1 \\1 & 1 & 1\end{array}\right|=2-1=1 \quad \Rightarrow z=\frac{1}{4} \\\end{array}$
For $\alpha=2 \Rightarrow$ unique solution
