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Chapter 6: Q15E (page 331)

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

Expert verified

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

Step by step solution

01

Use the least-square solution

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

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

02

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

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

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

03

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

The augmented matrix is:

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

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

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Most popular questions from this chapter

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.

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

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1. Show that the Cauchy-Schwarz inequality holds for \({\bf{x}} = \left( {{\bf{3}}, - {\bf{2}}} \right)\) and \({\bf{y}} = \left( { - {\bf{2}},{\bf{1}}} \right)\). (Suggestion: Study \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).)

In Exercises 7–10, let\[W\]be the subspace spanned by the\[{\bf{u}}\]’s, and write y as the sum of a vector in\[W\]and a vector orthogonal to\[W\].

10.\[y = \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\0\\1\\1\end{aligned}} \right]\],\[{{\bf{u}}_3} = \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

Let \(X\) be the design matrix in Example 2 corresponding to a least-square fit of parabola to data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\). Suppose \({x_1}\), \({x_2}\) and \({x_3}\) are distinct. Explain why there is only one parabola that best, in a least-square sense. (See Exercise 5.)
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