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Q40E

Expert-verified
Found in: Page 144

Linear Algebra With Applications

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

In Exercise 40 through 43, consider the problem of fitting a conic through${\mathbit{m}}$ given points${{\mathbit{P}}}_{{\mathbf{1}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{,}}{\mathbf{.}}{\mathbf{......}}{\mathbf{,}}{{\mathbit{P}}}_{{\mathbf{m}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{m}}\mathbf{,}{\mathbf{y}}_{\mathbf{m}}\mathbf{\right)}$ in the plane; see Exercise 53 through 62 in section 1.2. Recall that a conic is a curve in${{\mathbit{ℝ}}}^{{\mathbf{2}}}$ that can be described by an equation of the form , ${\mathbit{f}}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}{\mathbf{=}}{{\mathbit{c}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbit{c}}}_{{\mathbf{2}}}{\mathbit{x}}{\mathbf{+}}{{\mathbit{c}}}_{{\mathbf{3}}}{\mathbit{y}}{\mathbf{+}}{{\mathbit{c}}}_{{\mathbf{4}}}{{\mathbit{x}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbit{c}}}_{{\mathbf{5}}}{\mathbit{x}}{\mathbit{y}}{\mathbf{+}}{{\mathbit{c}}}_{{\mathbf{6}}}{{\mathbit{y}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{0}}$where at least one of the coefficients is non zero.40. Explain why fitting a conic through the points ${\mathbit{P}}_{\mathbf{1}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{1}}\mathbf{\right)}\mathbf{,}\mathbf{.}\mathbf{......}\mathbf{,}{\mathbit{P}}_{\mathbf{m}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{m}}\mathbf{,}{\mathbf{y}}_{\mathbf{m}}\mathbf{\right)}$amounts to finding the kernel of an${\mathbit{m}}{\mathbf{×}}{\mathbf{6}}$ matrix${\mathbit{A}}$ . Give the entries of the row of ${\mathbit{A}}$.Note that a one-dimensional subspace of the kernel of defines a unique conic, since the equations${\mathbit{f}}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}{\mathbf{=}}{\mathbf{0}}$ and ${\mathbit{k}}{\mathbit{f}}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}{\mathbf{=}}{\mathbf{0}}$describe the same conic.

Thus, the entries for$ith$ row are$\left[\begin{array}{cccccc}1& {x}_{i}& {y}_{i}& {x}_{i}^{2}& {x}_{i}{y}_{i}& {y}_{i}^{2}\end{array}\right]$ .

See the step by step solution

Step 1: Given information

A conic is a curve in${ℝ}^{2}$ that can be described by an equation of the form , $f\left(x,y\right)={c}_{1}+{c}_{2}x+{c}_{3}y+{c}_{4}{x}^{2}+{c}_{5}xy+{c}_{6}{y}^{2}=0$where at least one of the coefficients ${c}_{i}$is non-zero.

Step 2:Explanation

To fit a conic through the points${P}_{1}\left({x}_{1},{y}_{1}\right),.....,{P}_{m}\left({x}_{m},{y}_{m}\right)$ is equivalent to find the kernel of an $m{x}_{6}$of matrix and also these are solution to the equation.

$f\left(x,y\right)={c}_{1}+{c}_{2}x+{c}_{3}y+{c}_{4}{x}^{2}+{c}_{5}xy+{c}_{6}{y}^{2}=0$

Thus, the$ith$ row have entries as follows:

$\left[\begin{array}{cccccc}1& {x}_{i}& {y}_{i}& {x}_{i}^{2}& {x}_{i}{y}_{i}& {y}_{i}^{2}\end{array}\right]$

Hence, the entries for$ith$ row are$\left[\begin{array}{cccccc}1& {x}_{i}& {y}_{i}& {x}_{i}^{2}& {x}_{i}{y}_{i}& {y}_{i}^{2}\end{array}\right]$ .