Let the system of linear equations x+y+alpha | vedclass

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(C) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 1 & 1 & \alpha \ 3 & 1 & 1 \ 1 & 0 & 2 \end{vmatrix} = 1(2-0) - 1(6-1) +

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$2$

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(C) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 1 & 1 & \alpha \ 3 & 1 & 1 \ 1 & 0 & 2 \end{vmatrix} = 1(2-0) - 1(6-1) + \alpha(0-1) = 2 - 5 - \alpha = -\alpha - 3$.For a unique solution,$\Delta 
eq 0$,so $\alpha 
eq -3$.Using Cramer's rule:$x^{*} = \frac{\Delta_1}{\Delta} = \frac{\begin{vmatrix} 2 & 1 & \alpha \ 4 & 1 & 1 \ 1 & 0 & 2 \end{vmatrix}}{-(\alpha+3)} = \frac{2(2-0) - 1(8-1) + \alpha(0-1)}{-(\alpha+3)} = \frac{4-7-\alpha}{-(\alpha+3)} = \frac{-\alpha-3}{-(\alpha+3)} = 1$.$y^{*} = \frac{\Delta_2}{\Delta} = \frac{\begin{vmatrix} 1 & 2 & \alpha \ 3 & 4 & 1 \ 1 & 1 & 2 \end{vmatrix}}{-(\alpha+3)} = \frac{1(8-1) - 2(6-1) + \alpha(3-4)}{-(\alpha+3)} = \frac{7-10-\alpha}{-(\alpha+3)} = \frac{-\alpha-3}{-(\alpha+3)} = 1$.$z^{*} = \frac{\Delta_3}{\Delta} = \frac{\begin{vmatrix} 1 & 1 & 2 \ 3 & 1 & 4 \ 1 & 0 & 1 \end{vmatrix}}{-(\alpha+3)} = \frac{1(1-0) - 1(3-4) + 2(0-1)}{-(\alpha+3)} = \frac{1+1-2}{-(\alpha+3)} = 0$.Thus,$(x^{*}, y^{*}, z^{*}) = (1, 1, 0)$.The points are $(\alpha, 1), (1, \alpha)$ and $(1, -1)$.Since they are collinear,the area of the triangle formed by them is $0$:$\frac{1}{2} \begin{vmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & -1 & 1 \end{vmatrix} = 0 \Rightarrow \alpha(\alpha+1) - 1(1-1) + 1(-1-\alpha) = 0$.$\alpha^2 + \alpha - 1 - \alpha = 0 \Rightarrow \alpha^2 = 1 \Rightarrow \alpha = \pm 1$.Both values satisfy $\alpha 
eq -3$. The sum of absolute values is $|1| + |-1| = 1 + 1 = 2$.

Let the system of linear equations $x+y+\alpha z=2$,$3x+y+z=4$,and $x+2z=1$ have a unique solution $(x^{}, y^{}, z^{})$. If $(\alpha, x^{}), (y^{}, \alpha)$ and $(x^{}, -y^{*})$ are collinear points,then the sum of absolute values of all possible values of $\alpha$ is

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Let the system of linear equations $x+y+\alpha z=2$,$3x+y+z=4$,and $x+2z=1$ have a unique solution $(x^{*}, y^{*}, z^{*})$. If $(\alpha, x^{*}), (y^{*}, \alpha)$ and $(x^{*}, -y^{*})$ are collinear points,then the sum of absolute values of all possible values of $\alpha$ is