Question

In Exercises 15 and 16, use the factorization \(A = QR\) to find the least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

15. \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{3}}\\{\bf{2}}&{\bf{4}}\\{\bf{1}}&{\bf{1}}\end{aligned}} \right) = \left( {\begin{aligned}{{}{}}{\frac{{\bf{2}}}{{\bf{3}}}}&{ - \frac{{\bf{1}}}{{\bf{3}}}}\\{\frac{{\bf{2}}}{{\bf{3}}}}&{\frac{{\bf{2}}}{{\bf{3}}}}\\{\frac{{\bf{1}}}{{\bf{3}}}}&{ - \frac{{\bf{2}}}{{\bf{3}}}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{\bf{3}}&{\bf{5}}\\{\bf{0}}&{\bf{1}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{7}}\\{\bf{3}}\\{\bf{1}}\end{aligned}} \right)\)

Short answer

The least-square solution of \(A{\bf{x}} = {\bf{b}}\) is \(\left( {\begin{aligned}{{}{}}4\\{ - 1}\end{aligned}} \right)\).

Step-by-step solution

Use the least-square solution

Compare the given solution with the equation \(R{\bf{\hat x}} = {Q^T}{\bf{b}}\).

\({Q^T}{\bf{b}} = \left( {\begin{aligned}{{}{}}{\frac{2}{3}}&{\frac{2}{3}}&{\frac{1}{3}}\\{ - \frac{1}{3}}&{\frac{2}{3}}&{ - \frac{2}{3}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}7\\3\\1\end{aligned}} \right)\), \(R = \left( {\begin{aligned}{{}{}}3&5\\0&1\end{aligned}} \right)\)

Find the product \({Q^T}{\bf{b}}\)

The product \({Q^T}{\bf{b}}\) can be calculated as follows:

\(\begin{aligned}{}{Q^T}{\bf{b}} &= \left( {\begin{aligned}{{}{}}{\frac{2}{3}}&{\frac{2}{3}}&{\frac{1}{3}}\\{ - \frac{1}{3}}&{\frac{2}{3}}&{ - \frac{2}{3}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}7\\3\\1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}{\frac{{14}}{3} + \frac{6}{3} + \frac{1}{3}}\\{ - \frac{7}{3} + \frac{6}{3} - \frac{2}{3}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}7\\{ - 1}\end{aligned}} \right)\end{aligned}\)

Write the augmented matrix \(\left( {\begin{aligned}{{}{}}R&{{Q^T}{\bf{b}}}\end{aligned}} \right)\)

The augmented matrix is:

\(\begin{aligned}{}\left( {\begin{aligned}{{}{}}R&{{Q^T}{\bf{b}}}\end{aligned}} \right) &= \left( {\begin{aligned}{{}{}}3&5&7\\0&1&{ - 1}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}1&{\frac{5}{3}}&{\frac{7}{3}}\\0&1&{ - 1}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to \frac{{{R_1}}}{3}} \right)\\ &= \left( {\begin{aligned}{{}{}}1&0&4\\0&1&{ - 1}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to {R_1} - \frac{5}{3}{R_2}} \right)\end{aligned}\)

Thus, the least square solution of \(A{\bf{x}} = {\bf{b}}\) is \(\left( {\begin{aligned}{{}{}}4\\{ - 1}\end{aligned}} \right)\).