 Solution

Substitute the QR decomposition into the normal equations:The diagonal of $R$ is nonzero, so $R^T$ is invertible. Cancel it algebraically to obtainCompute the three inner products in $Q^{T}b$, then use back substitution in this upper triangular matrix. There is no need to form an inverse or explicitly construct $A^{T}A$. The orthogonal projection of $b$ onto the column space is $Q{Q}^{T}b$; its orthogonal residual proves that this solution minimizes $\Vert Ay-b{\Vert }_2$.