Compute the least-squares error associated with the least square solution found in Exercise 3.

From Exercise 3, the normal equation is:

\(\left( {\begin{aligned}{{}{}}6&6\\6&{42}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}{}}6\\{ - 6}\end{aligned}} \right)\)

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

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

The value of \(\left| {A{\bf{\hat x}} - {\bf{b}}} \right|\) can be calculated as:

\(\begin{aligned}{}\left| {A{\bf{\hat x}} - {\bf{b}}} \right| &= \sqrt {{{\left( 1 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 1 \right)}^2}} \\ &= \sqrt {1 + 9 + 9 + 1} \\ &= \sqrt {20} \\ &= 2\sqrt 5 \end{aligned}\)

Thus, the least square error is \(2\sqrt 5 \).