Question

In Exercises 5 and 6, describe all least squares solutions of the equation \(A{\bf{x}} = {\bf{b}}\).

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

Short answer

The solution is \({\bf{\hat x}} = \left( {\begin{aligned}{{}{}}5\\{ - 3}\\0\end{aligned}} \right) + {x_3}\left( {\begin{aligned}{{}{}}{ - 1}\\1\\1\end{aligned}} \right)\).

Step-by-step solution

Find the products \({A^T}A\) and \({A^T}{\bf{b}}\)

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

\(\begin{aligned}{}{A^T}A & = \left( {\begin{aligned}{{}{}}1&1&1&1\\1&1&0&0\\0&0&1&1\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1&1&0\\1&1&0\\1&0&1\\1&0&1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}4&2&2\\2&2&0\\2&0&2\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}{A^T}{\bf{b}}& = \left( {\begin{aligned}{{}{}}1&1&1&1\\1&1&0&0\\0&0&1&1\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1\\3\\8\\2\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{14}\\4\\{10}\end{aligned}} \right)\end{aligned}\)

Write the augmented matrix

The augmented matrix:

\(\begin{aligned}{}\left( {\begin{aligned}{{}{}}{{A^T}A}&{{A^T}{\bf{b}}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}{}}4&2&2&{14}\\2&2&0&4\\2&0&2&{10}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}1&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{7}{2}}\\2&2&0&4\\2&0&2&{10}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to \frac{{{R_1}}}{4}} \right)\\ & = \left( {\begin{aligned}{{}{}}1&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{7}{2}}\\0&1&{ - 1}&{ - 3}\\0&{ - 1}&1&3\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to {R_2} - 2{R_1}} \right)\\ & = \left( {\begin{aligned}{{}{}}1&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{7}{2}}\\0&1&{ - 1}&{ - 3}\\0&0&0&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_3} \to {R_3} + {R_2}} \right)\\ & = \left( {\begin{aligned}{{}{}}1&0&1&5\\0&1&{ - 1}&{ - 3}\\0&0&0&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to {R_1} - \frac{1}{2}{R_2}} \right)\end{aligned}\)

Find the solution \({\bf{\hat x}}\)

Using the augmented matrix the least square solution is:

\({\bf{\hat x}} = \left( {\begin{aligned}{{}{}}5\\{ - 3}\\0\end{aligned}} \right) + {x_3}\left( {\begin{aligned}{{}{}}{ - 1}\\1\\1\end{aligned}} \right)\)

Thus, the least square solution is \({\bf{\hat x}} = \left( {\begin{aligned}{{}{}}5\\{ - 3}\\0\end{aligned}} \right) + {x_3}\left( {\begin{aligned}{{}{}}{ - 1}\\1\\1\end{aligned}} \right)\).