Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Find the least square ${\overset{⇀}{x}}^{*}$ of the system $A \overset{⇀}{x} = b$ where $A = [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}]$ and $\overset{⇀}{b} = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]$ .

The solution $\overset{⇀}{x} = [\begin{matrix} - \frac{7}{6} \\ 1 \\ 0 \end{matrix}] + t [\begin{matrix} 1 \\ - 2 \\ 1 \end{matrix}]$ .

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Step by Step Solution

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TABLE OF CONTENTS

Step1: Explanation of the solution

Consider the system as follows.

$A \overset{⇀}{x} = b$ where $A = [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}]$ and $\overset{⇀}{b} = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]$ .

Consider the matrix A as follows.

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

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

$A^{T} A = {[\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}]}^{T} [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}] = [\begin{matrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}] = [\begin{matrix} 66 & 78 & 90 \\ 78 & 93 & 108 \\ 90 & 108 & 126 \end{matrix}] A^{T} A = [\begin{matrix} 66 & 78 & 90 \\ 78 & 93 & 108 \\ 90 & 108 & 126 \end{matrix}]$

The exact solution of the system is as follows.

$(A^{T} A) \overset{⇀}{x} = A^{T} \overset{⇀}{b} [\begin{matrix} 66 & 78 & 90 \\ 78 & 93 & 108 \\ 90 & 108 & 126 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}] [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}] [\begin{matrix} 66 & 78 & 90 \\ 78 & 93 & 108 \\ 90 & 108 & 126 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}] [\begin{matrix} 66 & 78 & 90 \\ 78 & 93 & 108 \\ 90 & 108 & 126 \end{matrix}] [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

After solving the matrix calculation as follows.

$[\begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix} \begin{matrix} - \frac{7}{6} \\ 1 \\ 0 \end{matrix}]$

Then write the equation as follows.

$x_{1} - x_{3} = - \frac{7}{6} x_{2} + 2 x_{3} = 1 x_{2} + 2 x_{3} = 1 x_{3} = t$

Simplify further as follows.

$[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} - \frac{7}{6} + t \\ 1 - 2 t \\ t \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] [\begin{matrix} - \frac{7}{6} + t \\ 1 \\ 0 \end{matrix}] + t [\begin{matrix} 1 \\ - 2 \\ 3 \end{matrix}]$

Hence, the least square for the system $(A^{T} A) \vec{x} = A^{T} \overset{⇀}{b}$ is $\overset{⇀}{x} = [\begin{matrix} - \frac{7}{6} \\ 1 \\ 0 \end{matrix}] + t [\begin{matrix} 1 \\ - 2 \\ 1 \end{matrix}]$ .

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Linear Algebra With Applications

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Short Answer

Find the least square ${\overset{⇀}{x}}^{*}$ of the system $A \overset{⇀}{x} = b$ where $A = [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}]$ and $\overset{⇀}{b} = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]$ .

Step by Step Solution

Step1: Explanation of the solution

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Find the least square ${\overset{⇀}{x}}^{*}$ of the system $A \overset{⇀}{x} = b$ where $A = [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}]$ and $\overset{⇀}{b} = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]$ .