Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1 / 8C (Vectors and Matrices)

Write $y=(y_1,y_2,y_3{)}^T$. From the coordinate formula for the cross product,If $y=c_1{e}_1$, this becomesThus $c_1\ne 0$ gives rank three and a zero-dimensional kernel, while $c_1=0$ gives rank one and kernel dimension two.

If $y\cdot e_1=0$, then $y_1=0$. When $y\ne 0$, the equations $Tx=0$ force $x_1=0$ and $y_3{x}_2-y_2{x}_3=0$, soWhen $y=0$, again $\mathrm{rank}T=1$ and $\dim \ker T=2$. These dimensions also agree with the rank-nullity theorem.

First computeThe vector $\ell =(1,-2,4{)}^T$ lies in the kernel of $(AB{)}^T$, since ${\ell }^{T}AB=0$. If $ABx=d$ is solvable, multiplying by ${\ell }^T$ gives the necessary compatibility conditionTherefore