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Q18E

Expert-verified
Found in: Page 400

### Linear Algebra With Applications

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

# Sketch the curves defined in Exercises 15 through 20. In each case, draw and label the principal axes, label the intercepts of the curve with the principal axes, and give the formula of the curve in the coordinate system defined by the principal axes.18.${\mathbf{9}}{{\mathbf{x}}}_{{\mathbf{1}}}^{{\mathbf{2}}}{\mathbf{·}}{\mathbf{4}}{{\mathbf{x}}}_{{\mathbf{1}}}{{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{6}}{{\mathbf{x}}}_{{\mathbf{2}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{1}}$

${\stackrel{\to }{v}}_{1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}-2\\ 1\end{array}\right] \mathrm{and} {\stackrel{\to }{v}}_{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ 2\end{array}\right]$

See the step by step solution

## Step 1: Given Information:

$9{x}_{1}^{2}-4{x}_{1}{x}_{2}+6{x}_{2}^{2}=1$

## Step 2: To Find the Eigen values:

$q\left({x}_{1},{x}_{2}\right)=9{x}_{1}^{2}-4{x}_{1}{x}_{2}+6{x}_{2}^{2}=1$

$=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]\left[\begin{array}{c}9{x}_{1}-2{x}_{2}\\ -2{x}_{1}+6{x}_{2}\end{array}\right]$

Split the term $-4{x}_{1}{x}_{2}$ equally between the two components. Therefore

$q\left(\stackrel{\to }{x}\right)=\stackrel{\to }{x}×A\stackrel{\to }{x}, \mathrm{where} A=\left[\begin{array}{cc}9& -2\\ -2& 6\end{array}\right]$

To determine the eigen values of the matrix A

$\mathrm{det}\left(\mathrm{A}-\mathrm{\lambda I}\right)=0\left|\begin{array}{cc}9-\mathrm{\lambda }& -2\\ -2& 6-\mathrm{\lambda }\end{array}\right|=0$

$\left(9-\lambda \right)\left(6-\lambda \right)-\left(-2\right)\left(-2\right)=0$

${\lambda }^{2}-15\lambda +50=0$

$\left(\lambda -10\right)\left(\lambda -5\right)=0$

The eigen values are ${\lambda }_{1}=10and{\lambda }_{2}=5$

To find the eigen vectors are ${\lambda }_{1}=10$

$\left(A-10I\right)\stackrel{~}{0}\left[\begin{array}{cc}9-10& -2\\ -2& 6-10\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

$\left[\begin{array}{cc}-1& -2\\ -2& -4\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

Apply Row operation${\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-2{\mathrm{R}}_{1}$

$\left[\begin{array}{cc}-1& -2\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

The first row implies ${x}_{1}=-2{x}_{2},{x}_{2}=1,{x}_{1}=-2,{\lambda }_{1}=10$

${\stackrel{\to }{u}}_{1}=\left[\begin{array}{c}-2\\ 1\end{array}\right]$

Orthonormal eigen basis simply by dividing the given eigen vector by its length

$\stackrel{\to }{{V}_{1}}=\frac{1}{||{\stackrel{\to }{u}}_{1}||}{\stackrel{\to }{u}}_{1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}-2\\ 1\end{array}\right]$

When ${\lambda }_{2}=5$

$\begin{array}{r}\left(A-5I\right)\overline{x}=0\left[\begin{array}{cc}9-5& -2\\ -2& 6-5\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0\\ \left[\begin{array}{lr}4& -2\\ -2& 1\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0\end{array}$

Apply row operation ${\mathrm{R}}_{2}\to 2{\mathrm{R}}_{2}+{\mathrm{R}}_{1}$

$\left[\begin{array}{cc}4& 2\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=0$

First row implies that ${x}_{1}=\frac{1}{2}{x}_{2},{x}_{2}=2{x}_{1}=1,{\lambda }_{2}=2$

$\stackrel{\to }{{u}_{2}}=\left[\begin{array}{c}1\\ 2\end{array}\right]$

Orthonormal eigen basis simply by dividing the given eigen vector by its length

$\stackrel{\to }{{v}_{2}}=\frac{1}{||{\stackrel{\to }{u}}_{2}||}{\stackrel{\to }{u}}_{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ 2\end{array}\right]$

$10{c}_{1}^{2}+5{c}_{2}^{2}=1$

${\stackrel{\to }{v}}_{1}=\frac{1}{\sqrt{5}}\left[\begin{array}{r}-2\\ 1\end{array}\right]\phantom{\rule{1em}{0ex}}\mathrm{and}\phantom{\rule{1em}{0ex}}{\stackrel{\to }{v}}_{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{l}1\\ 2\end{array}\right]$

## Step 3: To Find the graph of the axes:

The graph of the coordinate axes is