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Q26E

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Found in: Page 247

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Find the least square ${\stackrel{\mathbf{⇀}}{\mathbf{x}}}^{{\mathbf{*}}}$ of the system ${\mathbit{A}}\stackrel{\mathbf{⇀}}{\mathbf{x}}{\mathbf{=}}{\mathbit{b}}$ where ${\mathbit{A}}{\mathbf{=}}\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$ and $\stackrel{\mathbf{⇀}}{\mathbf{b}}{\mathbf{=}}\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$ .

The solution $\stackrel{⇀}{x}=\left[\begin{array}{c}-\frac{7}{6}\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]$.

See the step by step solution

## Step1: Explanation of the solution

Consider the system as follows.

$A\stackrel{⇀}{x}=b$where $A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$ and $\stackrel{⇀}{b}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$.

Consider the matrix A as follows.

$A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$

Now, find ${A}^{T}A$ as follows.

${A}^{T}A={\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]}^{T}\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccc}1& 4& 7\\ 2& 5& 8\\ 3& 6& 9\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right]\phantom{\rule{0ex}{0ex}}{A}^{T}A=\left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right]$

The exact solution of the system is as follows.

$\left({A}^{T}A\right)\stackrel{⇀}{x}={A}^{T}\stackrel{⇀}{b}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

After solving the matrix calculation as follows.

$\left[\overline{)\begin{array}{ccc}1& 0& -1\\ 0& 1& 2\\ 0& 0& 0\end{array}}\begin{array}{c}-\frac{7}{6}\\ 1\\ 0\end{array}\right]$

Then write the equation as follows.

${x}_{1}-{x}_{3}=-\frac{7}{6}\phantom{\rule{0ex}{0ex}}{x}_{2}+2{x}_{3}=1\phantom{\rule{0ex}{0ex}}{x}_{2}+2{x}_{3}=1\phantom{\rule{0ex}{0ex}}{x}_{3}=t$

Simplify further as follows.

$\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}-\frac{7}{6}+t\\ 1-2t\\ t\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]\left[\begin{array}{c}-\frac{7}{6}+t\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ -2\\ 3\end{array}\right]$

Hence, the least square for the system $\left({A}^{T}A\right)\stackrel{\to }{x}={A}^{T}\stackrel{⇀}{b}$ is $\stackrel{⇀}{x}=\left[\begin{array}{c}-\frac{7}{6}\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]$ .

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