Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 15

The homogeneous linear system has solution set $\ker A$, a vector subspace. If it contains only the zero vector, no two different solutions of the full system can exist: their difference would be a nonzero homogeneous solution.

A particular solution ${\mathbf{v}}_p$ to the full system exists exactly when $\mathbf{w}$ belongs to the image of a linear map $\mathrm{im}A$. If it exists, every solution isIndeed, subtracting $A{\mathbf{v}}_p=\mathbf{w}$ from $A\mathbf{v}=\mathbf{w}$ gives $A(\mathbf{v}-{\mathbf{v}}_p)=0$, and conversely adding a kernel vector preserves the right-hand side. Thus there are no solutions, one solution if the kernel is trivial, or infinitely many solutions if it is nontrivial. The last solution set is an affine subspace of dimension $\dim \ker A$, determined by the rank-nullity theorem. For a square nonsingular matrix there is one solution for every right-hand side.

Subtracting the third row from the first gives $z=-1$, and the third row gives $y=2-x$. The second row then reduces toFor $t\ne 1$, the unique solution isConsequently $s=1$ gives $\boxed{(x,y,z)=(1,1,-1)}$ for every $t\ne 1$, whereas $s=6$ gives

When $t=1$, the first and second left-hand sides coincide. If $s=1$, they are the same plane, and the remaining independent plane cuts it in the lineIf $s=6$, the two planes are distinct and parallel, so there is no solution. Geometrically, for $t\ne 1$ the three planes intersect at a single point; at $t=1$ the coincident-plane case leaves an entire line, while the parallel-plane case has empty intersection.