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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 a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

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

A simple curve that often makes a good model for the variable costs of a company, a function of the sales level \(x\), has the form \(y = {\beta _1}x + {\beta _2}{x^2} + {\beta _3}{x^3}\). There is no constant term because fixed costs are not included.

a. Give the design matrix and the parameter vector for the linear model that leads to a least-squares fit of the equation above, with data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\).

b. Find the least-squares curve of the form above to fit the data \(\left( {4,1.58} \right),\left( {6,2.08} \right),\left( {8,2.5} \right),\left( {10,2.8} \right),\left( {12,3.1} \right),\left( {14,3.4} \right),\left( {16,3.8} \right)\) and \(\left( {18,4.32} \right)\), with values in thousands. If possible, produce a graph that shows the data points and the graph of the cubic approximation.

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.)

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

  1. \(\left( {\begin{aligned}{{}{}}3\\0\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}8\\5\\{ - 6}\end{aligned}} \right)\)

Determine which pairs of vectors in Exercises 15-18 are orthogonal.

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)
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