Question

Let the system of linear equations
$x+y+\alpha z=2$
$3x+y+z=4$
$x+2z=1$
have a unique solution $\left(x^{\ast },y^{\ast },z^{\ast }\right)$. If $\left(\alpha ,x^{\ast }\right),\left(y^{\ast },\alpha \right)$ and $\left(x^{\ast },-y^{\ast }\right)$ 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)

$\Delta =\left|\begin{array}{ccc}1& 1& \alpha \\ 3& 1& 1\\ 1& 0& 2\end{array}\right|=-(\alpha +3)$
${\Delta }_1=\left|\begin{array}{ccc}2& 1& \alpha \\ 4& 1& 1\\ 1& 0& 2\end{array}\right|=-(3+\alpha )$
${\Delta }_2=\left|\begin{array}{ccc}1& 2& \alpha \\ 3& 4& 1\\ 1& 1& 2\end{array}\right|=-(\alpha +3)$
${\Delta }_3=\left|\begin{array}{ccc}1& 1& 2\\ 3& 1& 4\\ 1& 0& 1\end{array}\right|=0$
$\alpha \ne -3,x=1,y=1,z=0$,
Now points $(\alpha ,1),(1,\alpha )\&(1,-1)$ are collinear
$\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& -1& 1\end{array}\right|=0$
$\Rightarrow \alpha (\alpha +1)-1(1-1)+1(-1-\alpha )=0$
${\alpha }^2+\alpha -1-\alpha =0$
$\alpha =\pm 1$