Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 11 / b

Set $s=\mathbf{a}\cdot \mathbf{r}$. Taking the dot product of the vector equation with $\mathbf{a}$ givesHere $\mathbf{a}\cdot \mathbf{b}=5$ and $\mathbf{a}\cdot \mathbf{c}=6$, so $s=1$. Substitution then givesGeometrically the possible vector must lie on the line $\mathbf{c}-s\mathbf{b}$, and the scalar constraint picks one point. Equivalently the rank-one matrix update $I+\mathbf{b}{\mathbf{a}}^T$ is invertible because $1+\mathbf{a}\cdot \mathbf{b}=6\ne 0$. This is the nonsingular case of solvability of an affine rank-one vector equation.

The original PDF has $\mathbf{c}=(-2,1,1)$ here; the transcription loses its minus sign. Now $\mathbf{a}\cdot \mathbf{b}=-1$ and $\mathbf{a}\cdot \mathbf{c}=0$, so the scalar condition becomes $0=0$ and leaves $s$ arbitrary. ConsequentlyThe solution set is a straight line parallel to $\mathbf{b}$, equivalently the intersection of $y=z$ and $x+y=-1$.

For the geometric mechanism, let $L=I+\mathbf{b}{\mathbf{a}}^T$. Since $\mathbf{a}\cdot \mathbf{b}=-1$, $L^2=L$, $L\mathbf{b}=0$, and $\mathbf{a}\cdot L\mathbf{r}=0$. Thus $L$ is an oblique projection onto the plane ${\mathbf{a}}^{\perp }$ along $\mathbf{b}$. The specified $\mathbf{c}$ lies in this image plane, and adding any multiple of the kernel direction $\mathbf{b}$ leaves its image unchanged.

The same scalar consistency condition now has $\mathbf{a}\cdot \mathbf{c}=1$, so it reads $0=1$. Hence there is no solution.

Geometrically the image of $I+\mathbf{b}{\mathbf{a}}^T$ is still the plane ${\mathbf{a}}^{\perp }$, while this $\mathbf{c}$ is outside it. Thus the failure is inconsistency, rather than a unique solution with a singular denominator.