Consider the following system of equations: a | vedclass

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(B) The determinant of the coefficient matrix is $D = \begin{vmatrix} \alpha & 2 & 1 \ 2\alpha & 3 & 1 \ 3 & \alpha & 2 \end{vmatrix} = \alpha(6 - \

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It has no solution for $\alpha = -1$ and for all $\beta \in \mathbb{R}$.

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(B) The determinant of the coefficient matrix is $D = \begin{vmatrix} \alpha & 2 & 1 \ 2\alpha & 3 & 1 \ 3 & \alpha & 2 \end{vmatrix} = \alpha(6 - \alpha) - 2(4\alpha - 3) + 1(2\alpha^2 - 9) = 6\alpha - \alpha^2 - 8\alpha + 6 + 2\alpha^2 - 9 = \alpha^2 - 2\alpha - 3 = (\alpha - 3)(\alpha + 1)$.For $D = 0$,we have $\alpha = 3$ or $\alpha = -1$.If $\alpha = -1$,the system becomes $-x + 2y + z = 1$,$-2x + 3y + z = 1$,$3x - y + 2z = \beta$. Subtracting the first from the second gives $-x + y = 0$,so $x = y$. Substituting into the first gives $x + z = 1$,so $z = 1 - x$. Substituting into the third: $3x - x + 2(1 - x) = \beta \Rightarrow 2 = \beta$. Thus,if $\alpha = -1$ and $\beta 
eq 2$,there is no solution. If $\alpha = -1$ and $\beta = 2$,there are infinitely many solutions.If $\alpha = 3$,the system becomes $3x + 2y + z = 1$,$6x + 3y + z = 1$,$3x + 3y + 2z = \beta$. Subtracting the first from the second gives $3x + y = 0$,so $y = -3x$. Substituting into the first: $3x - 6x + z = 1 \Rightarrow z = 3x + 1$. Substituting into the third: $3x + 3(-3x) + 2(3x + 1) = \beta \Rightarrow 3x - 9x + 6x + 2 = \beta \Rightarrow 2 = \beta$. Thus,if $\alpha = 3$ and $\beta 
eq 2$,there is no solution.Option $B$ states it has no solution for $\alpha = -1$ and for all $\beta \in \mathbb{R}$,which is incorrect because it has infinitely many solutions when $\beta = 2$.

Consider the following system of equations: $\alpha x + 2y + z = 1$; $2\alpha x + 3y + z = 1$; $3x + \alpha y + 2z = \beta$. For some $\alpha, \beta \in \mathbb{R}$. Which of the following is $NOT$ correct?

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