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Chapter 6: Q16E (page 331)

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

Expert verified

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

Step by step solution

01

Use the least-square solution

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

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

02

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

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

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

03

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 \(\left( {\begin{aligned}{{}{}}{2.9}\\{.9}\end{aligned}} \right)\).

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Most popular questions from this chapter

A healthy child鈥檚 systolic blood pressure (in millimetres of mercury) and weight (in pounds) are approximately related by the equation

\({\beta _0} + {\beta _1}\ln w = p\)

Use the following experimental data to estimate the systolic blood pressure of healthy child weighing 100 pounds.

\(\begin{array} w&\\ & {44}&{61}&{81}&{113}&{131} \\ \hline {\ln w}&\\vline & {3.78}&{4.11}&{4.39}&{4.73}&{4.88} \\ \hline p&\\vline & {91}&{98}&{103}&{110}&{112} \end{array}\)

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set. Verify the following equality by induction, beginning with \(p = 2\). If \({\bf{x}} = {c_1}{{\bf{v}}_1} + \ldots + {c_p}{{\bf{v}}_p}\), then

\({\left\| {\bf{x}} \right\|^2} = {\left| {{c_1}} \right|^2} + {\left| {{c_2}} \right|^2} + \ldots + {\left| {{c_p}} \right|^2}\)

In Exercises 13 and 14, the columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that \(A = QR\). Check your work.

13. \(A = \left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right),{\rm{ }}Q = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{ - \frac{1}{6}}\\{\frac{1}{6}}&{\frac{5}{6}}\\{ - \frac{3}{6}}&{\frac{1}{6}}\\{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\)

In Exercises 11 and 12, find the closest point to \[{\bf{y}}\] in the subspace \[W\] spanned by \[{{\bf{v}}_1}\], and \[{{\bf{v}}_2}\].

12. \[y = \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right]\], \[{{\bf{v}}_1} = \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\], \[{{\bf{v}}_2} = \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\]

In Exercises 1-4, find the equation \(y = {\beta _0} + {\beta _1}x\) of the least-square line that best fits the given data points.

4. \(\left( {2,3} \right),\left( {3,2} \right),\left( {5,1} \right),\left( {6,0} \right)\)
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