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

The coefficient matrix, unknown vector and right-hand side are

Apply the determinant-preserving row operation $R_3\leftarrow R_3+3R_2$. The new third row is $(0,0,9-3\alpha )$, so expansion of the determinant along it givesThereforeThe parameter $\beta$ does not affect the determinant.

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

Call the three planes $P_1,P_2,P_3$ in their given order.

For $\alpha \ne 3,-1/2$, the nonzero determinant makes their normal vectors linearly independent. The three planes meet at the single point found in part (c)(i), for every $\beta$.

For $\alpha =3$, $P_2$ and $P_3$ coincide. Their common plane and $P_1$ have nonparallel normal vectors, since their first two coefficients are $(1,-1)$ and $(1,6)$. Thus they intersect in the straight line from part (c)(ii), for every $\beta$.

For $\alpha =-1/2$, write the three plane equations asThe last two meet on $z=0$, $x-y=-1$, which never lies on $P_1$. If $\beta =3$, $P_1$ and $P_2$ are distinct parallel planes, while $P_3$ cuts both. If $\beta =-1/2$, $P_1$ and $P_3$ are distinct parallel planes, while $P_2$ cuts both. For every other $\beta$, each pair of planes intersects in a straight line; these three pairwise intersection lines are distinct and parallel to $(1,1,0{)}^T$, with no common point. These exhaust the geometric cases.