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Chapter 6: Q3E (page 331)

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

Short Answer

Expert verified

(a) \(\left( {\begin{aligned}{{}{}}6&6\\6&{42}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}{}}6\\{ - 6}\end{aligned}} \right)\)

(b)\(\left( {\begin{aligned}{{}{}}{\frac{4}{3}}\\{ - \frac{1}{3}}\end{aligned}} \right)\)

Step by step solution

01

(a) Step 1: Find the products \({A^T}A\) and \({A^T}{\bf{b}}\)

Find the product \({A^T}A\).

\(\begin{aligned}{}{A^T}A & = \left( {\begin{aligned}{{}{}}1&{ - 1}&0&2\\{ - 2}&2&3&5\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1&{ - 2}\\{ - 1}&2\\0&3\\2&5\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}6&6\\6&{42}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}{A^T}{\bf{b}} & = \left( {\begin{aligned}{{}{}}1&{ - 1}&0&2\\{ - 2}&2&3&5\end{aligned}} \right)\left( {\begin{aligned}{{}{}}3\\1\\{ - 4}\\2\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}6\\{ - 6}\end{aligned}} \right)\end{aligned}\)

02

Find the solution by constructing the normal equations

The normal equations can be written as:

\(\begin{aligned}{}\left( {{A^T}A} \right){\bf{x}} & = {A^T}{\bf{b}}\\\left( {\begin{aligned}{{}{}}6&6\\6&{42}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}{}}6\\{ - 6}\end{aligned}} \right)\end{aligned}\)

03

(b) Step 3: Find the component \({\bf{\hat x}}\)

The component \({\bf{\hat x}}\) can be calculated as:

\(\begin{aligned}{}{\bf{\hat x}} & = {\left( {{A^T}A} \right)^{ - 1}}\left( {{A^T}{\bf{b}}} \right)\\ & = {\left( {\begin{aligned}{{}{}}6&6\\6&{42}\end{aligned}} \right)^{ - 1}}\left( {\begin{aligned}{{}{}}6\\{ - 6}\end{aligned}} \right)\\ & = \frac{1}{{216}}\left( {\begin{aligned}{{}{}}{288}\\{ - 72}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{\frac{4}{3}}\\{ - \frac{1}{3}}\end{aligned}} \right)\end{aligned}\)

The \({\bf{\hat x}}\) component is \(\left( {\begin{aligned}{{}{}}{\frac{4}{3}}\\{ - \frac{1}{3}}\end{aligned}} \right)\).

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Most popular questions from this chapter

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\)may be written in the form

\(\begin{aligned}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

16. Use a matrix inverse to solve the system of equations in (7) and thereby obtain formulas for \({\hat \beta _0}\) , and that appear in many statistics texts.

Let \(X\) be the design matrix used to find the least square line of fit data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\). Use a theorem in Section 6.5 to show that the normal equations have a unique solution if and only if the data include at least two data points with different \(x\)-coordinates.

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, determine which sets of vectors are orthogonal.

\(\left[ {\begin{align}{ 2}\\{ - 7}\\{-1}\end{align}} \right]\), \(\left[ {\begin{align}{ - 6}\\{ - 3}\\9\end{align}} \right]\), \(\left[ {\begin{align}{ 3}\\{ 1}\\{-1}\end{align}} \right]\)

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.

6. \(\left( {\begin{aligned}{{}}3\\{ - 1}\\2\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}}{ - 5}\\9\\{ - 9}\\3\end{aligned}} \right)\)
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