Past exam of the mathematical methods I course of the University of Oxford / 2017 resit / 5

Gaussian elimination of the first two equations gives $x=2y-1$ and $z=3-4y$. The third then reduces to $(\alpha -2)y=\beta -3$. HenceIn the infinite family $y=t$, $x=2t-1$ and $z=3-4t$, for $t\in \mathbb{R}$. This makes the distinction between an inconsistent linear system and a free parameter explicit; its coefficient matrix has determinant $4-2\alpha$.

With $\alpha =\beta =6$, the eliminated equation gives $4y=3$. Back substitution yieldsThe three left-hand sides are respectively $4$, $2$ and $6$, verifying the solution in the original linear system. Its determinant is $-8\ne 0$, so this solution is unique.