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Let the system of linear equations

$\begin{aligned} &x+y+\alpha z=2 \\ &3 x+y+z=4 \\ &x+2 z=1 \end{aligned}$

Have a unique solution $\mathrm{\left (x^{\ast},y^{\ast},z^{\ast} \right )}$ If $\mathrm{\left (\alpha ,x^{\ast} \right )}$ , $\mathrm{\left (y^{\ast} ,\alpha \right )}$ and $\mathrm{\left (x^{\ast},-y^{\ast} \right )}$ are collinear points, then the sum of absolute values of all possible values of $\mathrm{\alpha }$ is
Option: 1
4

Option: 2
3

Option: 3
2

Option: 4
1

Answers (1)

best_answer

$\mathrm{x+y+\alpha z =2 }\\$

$\mathrm {3 x+y+z =4} \\$

$\mathrm{x+2 z =1}$

For unique solution:

$D=\left|\begin{array}{lll} 1 & 1 & \alpha \\ 3 & 1 & 1 \\ 1 & 0 & 2 \end{array}\right| \neq 0$

$\begin{gathered} 2-5-\alpha \neq 0 \\ \alpha \neq-3 \end{gathered}$

$D_{1}=\left|\begin{array}{lll} 2 & 1 & \alpha \\ 4 & 1 & 1 \\ 1 & 0 & 2 \end{array}\right|=-\alpha-3$

$D_{2}=\left|\begin{array}{lll} 1 & 2 & \alpha \\ 3 & 4 & 1 \\ 1 & 1 & 2 \end{array}\right|=-\alpha-3$

$D_{3}=\begin{vmatrix} 1 & 1 &2 \\ 3 &1 &4 \\ 1 & 0 &1 \end{vmatrix}=0$

Hunce $\alpha \neq-3, \quad x=1=y, z=0$

Non

$\begin{gathered} \left|\begin{array}{ccc} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & -1 & 1 \end{array}\right|=0 \\ \alpha^{2}=1 \\ \alpha=\pm 1 \\ \end{gathered}$

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