Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 17 / c

Adding three times the second linear equation to the third gives $3(3-\alpha )z=0$. Since $\alpha \ne 3$, it follows that $z=0$. The second linear equation then gives $y=x+1$, and the first becomes $(1+2\alpha )x=1-2\alpha$. Thus the unique solution isThe assumption $\alpha \ne -1/2$ makes the denominator nonzero; $\beta$ is unrestricted.

When $\alpha =3$, the third linear equation is $-3$ times the second, so it adds no constraint. Take $z=t$ freely. The remaining linear equations are $x-y=-1-3t$ and $x+6y=1-\beta t$. Solving gives the straight line of solutionsThis exists for every real $\beta$.

Here $\alpha \ne 3$, so the same elimination forces $z=0$. The first two linear equations would then require $x-y=1$ and $x-y=-1$ simultaneously. Thus