Question

# Find a least squares solution of Ax=b by constructing and solving the normal equations. A=[(3,1),(1,1),(1,4)], b[(1),(1),(1)] bar(x)=?

Equations
Find a least squares solution of Ax=b by constructing and solving the normal equations.
$$\displaystyle{A}={\left[\begin{array}{cc} {3}&{1}\\{1}&{1}\\{1}&{4}\end{array}\right]},{b}{\left[\begin{array}{c} {1}\\{1}\\{1}\end{array}\right]}$$
$$\displaystyle\overline{{{x}}}=$$?

2020-10-19
Step 1
We have to find the least square solution of Ax = B, by constructing the normal equations, where
$$\displaystyle{A}={\left[\begin{array}{cc} {3}&{1}\\{1}&{1}\\{1}&{4}\end{array}\right]},{b}{\left[\begin{array}{c} {1}\\{1}\\{1}\end{array}\right]}$$
The set of solutions of the non-empty solutions is given by
$$\displaystyle{A}^{{T}}{A}{x}={A}^{{T}}$$ b.To solve this normal equations, we first compute the relevant matrices.
$$\displaystyle{A}^{{T}}{A}={\left[\begin{array}{ccc} {3}&{1}&{1}\\{1}&{1}&{4}\end{array}\right]}{\left[\begin{array}{cc} {3}&{1}\\{1}&{1}\\{1}&{4}\end{array}\right]}={\left[\begin{array}{cc} {11}&{8}\\{8}&{18}\end{array}\right]}$$
$$\displaystyle{A}^{{T}}{b}={\left[\begin{array}{ccc} {3}&{1}&{1}\\{1}&{1}&{4}\end{array}\right]}{\left[\begin{array}{c} {1}\\{1}\\{1}\end{array}\right]}={\left[\begin{array}{c} {5}\\{6}\end{array}\right]}$$
Step 2
Now, we need to solve $$\displaystyle{\left[\begin{array}{cc} {11}&{8}\\{8}&{18}\end{array}\right]}{x}={\left[\begin{array}{c} {5}\\{6}\end{array}\right]}$$.
The augmented matrix is given by
$$\displaystyle{\left[\begin{array}{ccc} {11}&{8}&{5}\\{8}&{18}&{6}\end{array}\right]}\rightarrow{\left[\begin{array}{ccc} -{3}&{10}&{1}\\{8}&{18}&{6}\end{array}\right]}$$
$$\displaystyle\rightarrow{\left[\begin{array}{ccc} {1}&{10}&-{3}\\{6}&{18}&{8}\end{array}\right]}$$
$$\displaystyle\rightarrow{\left[\begin{array}{ccc} {1}&{10}&-{3}\\{0}&-{42}&-{10}\end{array}\right]}$$
$$\displaystyle\rightarrow{\left[\begin{array}{ccc} {1}&{10}&-{3}\\{0}&{21}&{5}\end{array}\right]}$$
Step 3
From the final matrix, we get the following equations $$\displaystyle{x}_{{1}}+{10}{x}_{{2}}=-{3}$$
$$\displaystyle{21}{x}_{{2}}={5}$$
$$\displaystyle\Rightarrow{x}_{{2}}=\frac{{5}}{{21}},{x}_{{1}}=-\frac{{113}}{{21}}$$
$$\displaystyle\Rightarrow{x}={\left[\begin{array}{c} \frac{{-{113}}}{{21}}\\\frac{{5}}{{21}}\end{array}\right]}$$