Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof resit / 6 / b / iii

The supplied factors give $Q^{T}b=(0,2{)}^T$. Back substitution in $Rx=Q^{T}b$ yields $2y=2$ and $10x+4y=0$, soMultiplying by the original matrix gives $(1,1,1,1{)}^T$, so this least-squares minimizer has residual zero: this particular overdetermined system is exactly consistent. Full column rank still guarantees uniqueness of the minimizer.