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Linear Algebra and its Applications
Found in: Page 331
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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 \({\bf{\hat x}}\).

2. \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{0}}\\{\bf{2}} {\bf{3}}\end{aligned}} \right)\), \(b = \left( {\begin{aligned}{{}{}}{ - {\bf{5}}}\\{\bf{8}}\\{\bf{1}}\end{aligned}} \right)\)

(a) \(\left( {\begin{aligned}{{}{}}{12}&8\\8&{10}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}{}}{ - 24}\\{ - 2}\end{aligned}} \right)\)

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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

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

\(\begin{aligned}{}{A^T}A & = \left( {\begin{aligned}{{}{}}2&{ - 2}&2\\1&0&3\end{aligned}} \right)\left( {\begin{aligned}{{}{}}2&1\\{ - 2}&0\\2&3\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{12}&8\\8&{10}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}{A^T}{\bf{b}} & = \left( {\begin{aligned}{{}{}}2&{ - 2}&2\\1&0&3\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{ - 5}\\8\\1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{ - 24}\\{ - 2}\end{aligned}} \right)\end{aligned}\)

Step 2: Find the solution by constructing the normal equations

The normal equations can be written as:

\(\begin{aligned}{}\left( {{A^T}A} \right){\bf{x}} & = {A^T}{\bf{b}}\\\left( {\begin{aligned}{{}{}}{12}&8\\8&{10}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}{}}{ - 24}\\{ - 2}\end{aligned}} \right)\end{aligned}\)

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

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

\(\begin{aligned}{}{\bf{\hat x}} & = {\left( {{A^T}A} \right)^{ - 1}}\left( {{A^T}{\bf{b}}} \right)\\ & = {\left( {\begin{aligned}{{}{}}{12}&8\\8&{10}\end{aligned}} \right)^{ - 1}}\left( {\begin{aligned}{{}{}}{ - 24}\\{ - 2}\end{aligned}} \right)\\ & = \frac{1}{{56}}\left( {\begin{aligned}{{}{}}{ - 224}\\{168}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{ - 4}\\3\end{aligned}} \right)\end{aligned}\)

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

Most popular questions for Math Textbooks

A healthy child’s 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.

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

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.

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

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

In exercises 1-6, determine which sets of vectors are orthogonal.

  1. \(\left[ {\begin{array}{*{20}{c}}{ - 1}\\4\\{ - 3}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}5\\2\\1\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\{ - 4}\\{ - 7}\end{array}} \right]\)

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.

5. \(\left( {\begin{aligned}{{}{}}1\\{ - 4}\\0\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}7\\{ - 7}\\{ - 4}\\1\end{aligned}} \right)\)
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