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Chapter 6: Q-6-19SE (page 331)

In Exercises 17鈥�20, solve\(Ax = b\) and \(A\left( {\Delta x} \right) = \Delta b\) and show that the inequality (2) holds in each case. (See the discussion of ill-conditionedmatrices in Exercises 41鈥�43 in Section 2.3.)

\(19.\,\,A = \left[ {\begin{array}{*{20}{c}}7&{ - 6}&{ - 4}&1\\{ - 5}&1&0&{ - 2}\\{10}&{11}&7&{ - 3}\\{19}&9&7&1\end{array}} \right],\,\,b = \left[ {\begin{array}{*{20}{c}}{.100}\\{2.888}\\{ - 1.404}\\{1.462}\end{array}} \right],\,\,\Delta b = \left[ {\begin{array}{*{20}{c}}{.49}\\{ - 1.28}\\{5.78}\\{8.04}\end{array}} \right]\)

Short Answer

Expert verified

The condition \(\frac{{\parallel \Delta x\parallel }}{{\parallel x\parallel }} \le cond(A) \cdot \frac{{\parallel \Delta b\parallel }}{{\parallel b\parallel }}\)is satisfied.

Step by step solution

01

The condition for solving matrix 

Enter the matrices in MATLAB

\(\begin{array}{c}A = \left[ {7{\rm{ }} - 6{\rm{ }} - 4{\rm{ }}1;{\rm{ }} - 5{\rm{ }}1{\rm{ }}0{\rm{ }} - 2;{\rm{ }}10{\rm{ }}11{\rm{ }}7{\rm{ }} - 3;{\rm{ }}19{\rm{ }}9{\rm{ }}7{\rm{ }}1} \right]\\b = \left[ {0.1;{\rm{ }}2.888;{\rm{ }} - 1.404;{\rm{ }}1.462} \right]\\delta\,\,b = {10^{( - 4)}}*[0.49; - 1.28;5.78;8.04]\end{array}\)

02

Solve the equation matrix

Solve the system of equation

\(\)

\(\begin{array}{c}Ax = b\\x = {A^{ - 1}}b\\x{\rm{ }} = \left[ { - 8.642,\,\, - 1730.054,\,\,2368.716,\,\, - 844.866} \right]\end{array}\)

Solve the system of equation as:

\(\begin{array}{c}A\left( {delta\,\,x} \right) = delta\,\,b\\delta\,\,x = {A^{ - 1}}\left( {delta\,\,b} \right)\\delta\,\,x = {10^{ - 4}}\left[ \begin{array}{c}0.3\\1.4\\ - 1.55\\0.59\end{array} \right]\end{array}\)

03

Find norm and condition number

\(\begin{array}{l}norm\left( {delta\,\,x} \right)/norm\left( x \right)\\ans{\rm{ }} = 7.1778e - 08\\norm\left( {delta\,\,b} \right)/norm(b)\\ans{\rm{ }} = 2.8320e - 04\end{array}\)

\(\begin{array}{l}ans{\rm{ }} = \\1.5479e - 04\\Condition{\rm{ }}Number{\rm{ }}of{\rm{ }}A:\\cond\left( A \right)\\\end{array}\)

Condition Number of A:

\(\begin{array}{l}Cond\left( A \right):\\ans{\rm{ }} = 2.3683e + 04\end{array}\)

04

Find ratio condition

Compute the ratio condition\((A) \cdot \frac{{\parallel \Delta b\parallel }}{{\parallel b\parallel }}\)

\(cond\left( A \right)*{\rm{ }}norm\left( {delta\,\,b} \right)/norm\left( b \right)\)

\(ans{\rm{ }} = 6.7071\)\(\)

So, the condition\(\frac{{\parallel \Delta x\parallel }}{{\parallel x\parallel }} \le cond(A) \cdot \frac{{\parallel \Delta b\parallel }}{{\parallel b\parallel }}\)is satisfied.

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Most popular questions from this chapter

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

7.\[y = \left[ {\begin{aligned}1\\3\\5\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\3\\{ - 2}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}5\\1\\4\end{aligned}} \right]\]

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 \(\overline x = \frac{1}{n}\left( {{x_1} + \cdots + {x_n}} \right)\), and \(\overline y = \frac{1}{n}\left( {{y_1} + \cdots + {y_n}} \right)\). Show that the least-squares line for the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) must pass through \(\left( {\overline x ,\overline y } \right)\). That is, show that \(\overline x \) and \(\overline y \) satisfies the linear equation \(\overline y = {\hat \beta _0} + {\hat \beta _1}\overline x \). (Hint: Derive this equation from the vector equation \({\bf{y}} = X{\bf{\hat \beta }} + \in \). Denote the first column of \(X\) by 1. Use the fact that the residual vector \( \in \) is orthogonal to the column space of \(X\) and hence is orthogonal to 1.)

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

3.\[y = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{4}}\\{\bf{3}}\end{aligned}} \right]\],\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{0}}\end{aligned}} \right]\)

Find a \(QR\) factorization of the matrix in Exercise 12.
See all solutions

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