 Solution

Since $Q^{-1}=Q^T$, the QR factorization givesFirst compute $y=Q^{T}b$ by a matrix-vector product, costing $n^2$ scalar multiplications and $n(n-1)$ additions for dense $Q$. Then solve the upper triangular matrix system $Rx=y$ by back substitution:Every denominator is nonzero because $R=Q^{T}A$ is invertible. The back substitution costs $n(n-1)$ multiplications and subtractions together, plus $n$ divisions, for $n^2$ arithmetic operations. Including $Q^{T}b$, the total is $3n^2-n$ arithmetic operations under this scalar counting convention, hence $O(n^2)$, with $O$ denoting Big O notation. The already supplied QR factorization is reused; no matrix inverse needs to be computed.