Question

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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

$\Delta=\begin{vmatrix}1 & 1 & \alpha \\ 3 & 1 & 1 \\ 1 & 0 & 2\end{vmatrix}=-(\alpha+3)$
$\Delta_{1}=\begin{vmatrix}2 & 1 & \alpha \\ 4 & 1 & 1 \\ 1 & 0 & 2\end{vmatrix}=-(3+\alpha)$
$\Delta_{2}=\begin{vmatrix}1 & 2 & \alpha \\ 3 & 4 & 1 \\ 1 & 1 & 2\end{vmatrix}=-(\alpha+3)$
$\Delta_{3}=\begin{vmatrix}1 & 1 & 2 \\ 3 & 1 & 4 \\ 1 & 0 & 1\end{vmatrix}=0$
$\alpha \neq-3, x=1, y =1, z =0$,
Now points $(\alpha, 1),(1, \alpha) \&(1,-1)$ are collinear
$\begin{vmatrix}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & -1 & 1\end{vmatrix}=0$
$\Rightarrow \alpha(\alpha+1)-1(1-1)+1(-1-\alpha)=0$
$\alpha^{2}+\alpha-1-\alpha=0$
$\alpha=\pm 1$