Past exam of the mathematical methods I course of the University of Oxford / 2013 resit / 5

Subtracting the first equation from the third gives $4x-4z=6$, so $z=x-3/2$. The second equation then becomes $(\alpha -2)x=\beta -3$. The third fixes $y=7(1-x)/5$. Therefore the three possibilities areFor the first case the coefficient determinant is $20(\alpha -2)\ne 0$, confirming uniqueness. At $\alpha =2$ its matrix rank is two. The augmented matrix has the same matrix rank only for $\beta =3$, giving one free variable in a linear system; otherwise the linear system is inconsistent.