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

uneskovogl5 2021-11-20 Answered
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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Expert Answer

Sevensis1977
Answered 2021-11-21 Author has 152 answers

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
\[\hat{x}=\begin{bmatrix}-4\\3 \end{bmatrix}\]

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