Question

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

16. \(A = \left( {\begin{aligned}{{}{}}1&{ - 1}\\1&4\\1&{ - 1}\\1&4\end{aligned}} \right) = \left( {\begin{aligned}{{}{}}{\frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}2&3\\0&5\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{ - 1}\\6\\5\\7\end{aligned}} \right)\)

Short answer

The least-square solution of \(A{\bf{x}} = {\bf{b}}\) is \(\left( {\begin{aligned}{{}{}}{2.9}\\{.9}\end{aligned}} \right)\).

Step-by-step solution

Use the least-square solution

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

\({Q^T}{\bf{b}} = \left( {\begin{aligned}{{}{}}{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{ - 1}\\6\\5\\7\end{aligned}} \right)\), \(R = \left( {\begin{aligned}{{}{}}2&3\\0&5\end{aligned}} \right)\)

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

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

\(\begin{aligned}{}{Q^T}{\bf{b}} &= \left( {\begin{aligned}{{}{}}{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{ - 1}\\6\\5\\7\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}{ - \frac{1}{2} + \frac{6}{2} + \frac{5}{2} + \frac{7}{3}}\\{\frac{1}{2} + \frac{6}{2} - \frac{5}{2} + \frac{7}{2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}{\frac{{17}}{2}}\\{\frac{9}{2}}\end{aligned}} \right)\end{aligned}\)

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

The augmented matrix is:

\(\begin{aligned}{}\left( {\begin{aligned}{{}{}}R&{{Q^T}{\bf{b}}}\end{aligned}} \right) &= \left( {\begin{aligned}{{}{}}2&3&{\frac{{19}}{2}}\\0&5&{\frac{9}{2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}1&{\frac{3}{2}}&{\frac{{19}}{4}}\\0&1&{\frac{9}{{10}}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to \frac{{{R_1}}}{2},{R_2} \to \frac{{{R_2}}}{5}} \right)\\ &= \left( {\begin{aligned}{{}{}}1&0&{\frac{{29}}{{10}}}\\0&1&{\frac{9}{{10}}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to {R_1} - \frac{3}{2}{R_2}} \right)\end{aligned}\)

Thus, the least square solution of \(A{\bf{x}} = {\bf{b}}\) is \(\left( {\begin{aligned}{{}{}}{2.9}\\{.9}\end{aligned}} \right)\).