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

3. \(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{ - {\bf{2}}}\\{ - {\bf{1}}}&{\bf{2}}\\{\bf{0}}&{\bf{3}}\\{\bf{2}}&{\bf{5}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{3}}\\{\bf{1}}\\{ - {\bf{4}}}\\{\bf{2}}\end{aligned}} \right)\)

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

\(\begin{aligned}{}{A^T}A & = \left( {\begin{aligned}{{}{}}1&{ - 1}&0&2\\{ - 2}&2&3&5\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1&{ - 2}\\{ - 1}&2\\0&3\\2&5\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}6&6\\6&{42}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}{A^T}{\bf{b}} & = \left( {\begin{aligned}{{}{}}1&{ - 1}&0&2\\{ - 2}&2&3&5\end{aligned}} \right)\left( {\begin{aligned}{{}{}}3\\1\\{ - 4}\\2\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}6\\{ - 6}\end{aligned}} \right)\end{aligned}\)

The normal equations can be written as:

\(\begin{aligned}{}\left( {{A^T}A} \right){\bf{x}} & = {A^T}{\bf{b}}\\\left( {\begin{aligned}{{}{}}6&6\\6&{42}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}{}}6\\{ - 6}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}{\bf{\hat x}} & = {\left( {{A^T}A} \right)^{ - 1}}\left( {{A^T}{\bf{b}}} \right)\\ & = {\left( {\begin{aligned}{{}{}}6&6\\6&{42}\end{aligned}} \right)^{ - 1}}\left( {\begin{aligned}{{}{}}6\\{ - 6}\end{aligned}} \right)\\ & = \frac{1}{{216}}\left( {\begin{aligned}{{}{}}{288}\\{ - 72}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{\frac{4}{3}}\\{ - \frac{1}{3}}\end{aligned}} \right)\end{aligned}\)

The \({\bf{\hat x}}\) component is \(\left( {\begin{aligned}{{}{}}{\frac{4}{3}}\\{ - \frac{1}{3}}\end{aligned}} \right)\).