Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 5 - Inner Product Spaces - 5.5 Chapter Review - Additional Problems - Page 377: 17

Answer

$y=2x+7$

Work Step by Step

The matrices can be formed as: $A=\begin{bmatrix} -4 & 1 \\ -3& 1 \\ -2 & 1\\ 0 &1 \end{bmatrix}$ $x=\begin{bmatrix} a \\ b \end{bmatrix}$ $b=\begin{bmatrix} -1\\ 1\\ 3\\ 7 \end{bmatrix}$ Apply matrices to the least square solution: $x_0=(A^TA)^{-1}A^Tb$ $=(\begin{bmatrix} -4& -3& -2 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} -4 & 1 \\ -3& 1 \\ -2& 1\\ 0&1 \\ \end{bmatrix})^{-1} \begin{bmatrix} -4& -3& -2 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix}\begin{bmatrix} -1\\ 1 \\ 3\\ 7 \end{bmatrix}$ $=\begin{bmatrix} 29 & -9\\ -9 & 4 \end{bmatrix}^{-1} \begin{bmatrix} -4& -3& -2 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} -1\\ 1 \\ 3\\ 7 \end{bmatrix}$ $=\frac{1}{35}\begin{bmatrix} 4 & 9\\ 9 & 29 \end{bmatrix} \begin{bmatrix} -4& -3& -2 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} -1\\ 1 \\ 3 \\ 7 \end{bmatrix}$ $=\frac{1}{35}\begin{bmatrix} -7& -3& 1 & 9\\ -7& 2 & 11 & 29\\ \end{bmatrix}\begin{bmatrix} -1\\ 1 \\ 3\\ 7 \end{bmatrix}$ $=\frac{1}{35}\begin{bmatrix} 70\\ 245 \end{bmatrix}$ $=\begin{bmatrix} 2\\ 7 \end{bmatrix}$ The equation of the least squares line associated with the given set of data points is $y=2x+7$
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