JEE Main · 2005 · 08 May 2005 · Offline · Matrices

The system has a unique solution when its coefficient determinant is non-zero, so the cases of interest are where it vanishes. Expanding, $\Delta =\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|=(\alpha -1{)}^2(\alpha +2)$, which is zero at $\alpha =1$ and $\alpha =-2$. At $\alpha =1$ all three equations reduce to $x+y+z=0$, giving infinitely many solutions. At $\alpha =-2$ the equations become mutually inconsistent, so the system fails to have a solution precisely when $\alpha =-2$.