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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

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

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