JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Matrices

The coefficient determinant is $\Delta =\alpha ({\alpha }^2-1)-(\alpha -1)+(1-\alpha )=(\alpha -1{)}^2(\alpha +2)$, which vanishes only at $\alpha =1$ and $\alpha =-2$. For $\alpha =-2,\beta =1$, adding the three equations yields $0=3$, so the system is inconsistent and has no solution. For $\alpha =1,\beta =1$, all three equations become $x+y+z=1$, giving infinitely many solutions. For $\alpha =2,\beta =1$, the determinant equals $4\ne 0$ and symmetry gives $x=y=z=\frac{1}{4}$, so $x+y+z=\frac{3}{4}$. For $\alpha =2,\beta =-1$, the determinant is again $4\ne 0$, so there is a unique solution; the assertion that this case has infinitely many solutions is the statement that is not correct.