 Solution

Let $a_1,a_2,a_3$ denote the columns. The Gram-Schmidt process begins with $\Vert a_1\Vert =2$, soThe second column has $r_{12}=q_1^{\mathsf{T}}{a}_2=4$ and residualThis residual has norm $2$, giving $q_2=\frac{1}{2}(-1,1,-1,1{)}^{\mathsf{T}}$ and $r_{22}=2$.

For the third column the orthogonal projections are $r_{13}=6$ and $r_{23}=4$. Subtracting them givesIts norm is $2$, so $q_3=\frac{1}{2}(1,1,-1,-1{)}^{\mathsf{T}}$ and $r_{33}=2$. Thus the requested thin QR decomposition isThe three displayed columns have pairwise zero inner products and unit norms; direct multiplication gives $QR=A$.