# Find a least-squares solution of Ax=b by (a) constructing the

Find a least-squares solution of Ax=b by (a) constructing the normal equations for $$\displaystyle\hat{{{x}}}$$ and (b) solving for $$\displaystyle\hat{{{x}}}$$
$A=\begin{bmatrix}2&1\\-2&0\\2&3\end{bmatrix},b=\begin{bmatrix}-5\\8\\1\end{bmatrix}$

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Sevensis1977

We know that each least square solution of $$x=b$$ satisfies the equation $$\displaystyle{A}^{{T}}{A}{x}={A}^{{T}}{b}$$. Therefore first of all find $$\displaystyle{A}^{{T}}{A},{A}^{{T}}{b}$$
$A^TA=\begin{bmatrix}2 & -2&2 \\1 &0&3 \end{bmatrix}\begin{bmatrix}2 & 1 \\-2 & 0\\2&3 \end{bmatrix}=\begin{bmatrix}12 & 8 \\8 &10 \end{bmatrix}$
$A^Tb=\begin{bmatrix}2 & -2&2 \\1 &0&3 \end{bmatrix}\begin{bmatrix}-5&8&1 \end{bmatrix}=\begin{bmatrix}-24\\-2\end{bmatrix}$
The equation $$\displaystyle{A}^{{T}}{A}{x}={A}^{{T}}{b}$$ becomes.
$\begin{bmatrix}12& 8 \\8& 10 \end{bmatrix}\begin{bmatrix}x_1 \\x_2\end{bmatrix}=\begin{bmatrix}-24\\-2 \end{bmatrix}$
The augmented matrix for $$\displaystyle{A}^{{T}}{A}{x}={A}^{{T}}{b}$$ to solve this sytem
$\begin{bmatrix}12&8&-24\\8&10&-2\end{bmatrix}$
Use row operation to solve this system
$\begin{bmatrix}12&8&-24\\8&10&-2\end{bmatrix}\sim\begin{bmatrix}1&0&-4\\0& 1&3 \end{bmatrix}$
The solution of the augmented matrix.
$$\displaystyle{x}_{{1}}=-{4},\ {x}_{{2}}={3}$$
The general least square solution for Ax=b
$\hat{x}=\begin{bmatrix}-4\\3 \end{bmatrix}$