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

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 \({A^T}{\bf{b}}\).

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

The normal equations can be written as:

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

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