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### 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

# 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 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).