Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 19D / c

First rotate rows $2,3$ with $c=4/5$, $s=3/5$. The second column pair $(4,3)$ becomes $(5,0)$ and the corresponding third column pair $(1,2)$ becomes $(2,1)$. Next rotate rows $3,4$ with $c=4/5$, $s=3/5$, sending the third column pair $(1,3/4)$ to $(5/4,0)$. Thus $R=G_{34}{G}_{23}A$ and $Q=G_{23}^T{G}_{34}^T$. One exact QR decomposition isThe construction proves $Q^{T}Q=I$, and direct multiplication checks $QR=A$.

The orthogonal matrix $Q$ preserves the Euclidean norm, soThe first three entries of $Rx-Q^{T}b$ can be set to zero by backward substitution in a triangular system, while its fourth entry is always $3$. Solving the triangular system gives $x_3=16/5$, $x_2=(35-2x_3)/5=143/25$, and $x_1=(98/25-x_2-x_3)/3=-5/3$. Thus the unique least-squares solution and residual norm areUniqueness follows from the nonzero diagonal entries of the top triangular block, equivalently the full column rank of $A$.