For the system of linear equations a x+y+z=1, | vedclass

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Question

$\left|\begin{array}{lll}\alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha\end{array}ight|=0$

$\alpha\left(\alpha^2-1i

Vedclass · Accepted Answer

It has infinitely many solutions if $\alpha=2$ and $\beta=-1$

Vedclass · Answer

$\left|\begin{array}{lll}\alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha\end{array}ight|=0$

$\alpha\left(\alpha^2-1ight)-1(\alpha-1)+1(1-\alpha)=0$

$\alpha^3-3 \alpha+2=0$

$\alpha^2(\alpha-1)+\alpha(\alpha-1)-2(\alpha-1)=0$

$(\alpha-1)\left(\alpha^2+\alpha-2ight)=0$

$\alpha=1, \alpha=-2,1$

$	ext { For } \alpha=1, \beta=1$

$\left.\begin{array}{l}x+y+z=1 \ x+y+z=b\end{array}ight\} 	ext { infinite solution }$

For $\alpha=2, \beta=1$

$\begin{array}{l}\Delta=4 \\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \1 & 2 & 1 \1 & 1 & 2\end{array}ight|=3-1-1 \quad \Rightarrow x =\frac{1}{4} \\Delta_2=\left|\begin{array}{lll}2 & 1 & 1 \1 & 1 & 1 \1 & 1 & 2\end{array}ight|=2-1=1\quad \Rightarrow y =\frac{1}{4} \\Delta_3=\left|\begin{array}{ccc}2 & 1 & 1 \1 & 2 & 1 \1 & 1 & 1\end{array}ight|=2-1=1 \quad \Rightarrow z=\frac{1}{4} \\end{array}$

For $\alpha=2 \Rightarrow$ unique solution

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 ?

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