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Q5.2-30E

Expert-verifiedFound in: Page 267

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Question: Let \(A = \left( {\begin{array}{*{20}{c}}{ - 6}&{28}&{21}\\4&{ - 15}&{ - 12}\\{ - 8}&a&{25}\end{array}} \right)\). For each value of \(a\) in the set \(\left\{ {32,31.9,31.8,32.1,32.2} \right\}\), compute the characteristic polynomial of \(A\) and the eigenvalues. In each case, create a graph of the characteristic polynomial \(p\left( t \right) = \det \left( {A - tI} \right)\) for \(0 \le t \le 3\). If possible, construct all graphs on one coordinate system. Describe how the graphs reveal the changes in the eigenvalues of \(a\) changes. **

Characteristic polynomial and eigenvalues are:

\[a\] | Characteristic polynomial | Eigenvalues |

\[31.8\] | \[ - .4 - 2.6t + 4{t^2} - {t^3}\] | \[3.1279,1, - .1279\] |

\[31.9\] | \[.8 - 3.8t + 4{t^2} - {t^3}\] | \[2.7042,1,.2958\] |

\[32.0\] | \[2 - 5t + 4{t^2} - {t^3}\] | \[2,1,1\] |

\[32.1\] | \[3.2 - 6.2t + 4{t^2} - {t^3}\] | \[1.5 \pm .9747i,1\] |

\[32.2\] | \[4.4 - 7.4t + 4{t^2} - {t^3}\] | \[1.5 \pm 1.4663i,1\] |

The graph of the characteristic polynomial is shown below:

Consider the matrix \(A = \left( {\begin{array}{*{20}{c}}{ - 6}&{28}&{21}\\4&{ - 15}&{ - 12}\\{ - 8}&a&{25}\end{array}} \right)\). Consider \(a = 32\)**.**

** **

Use the following command in the MATLAB to find the characteristic polynomial and eigenvalues of the matrix.

\[\begin{array}{l} > > {\rm{A}} = \left( {\begin{array}{*{20}{c}}{ - 6}&{28}&{21;}&4&{ - 15}&{ - 12;}\\{ - 8}&{32}&{25;}&{}&{}&{}\end{array}} \right);\\ > > {\rm{p}} = {\rm{poly}}\left( {\rm{A}} \right);\\ > > {\rm{eig}} = {\rm{eign}}\left( {\rm{A}} \right)\end{array}\]

So, the characteristic polynomial and eigenvalues of *A* is \(p = 2 - 5t + 4{t^2} - {t^3}\), \({\rm{eig}} = \left\{ {2,1,1} \right\}\).

Use the following command in MATLAB to find the characteristic polynomial and eigenvalues of the matrix.

\[\begin{array}{l} > > {\rm{A}} = \left( {\begin{array}{*{20}{c}}{ - 6}&{28}&{21;}&4&{ - 15}&{ - 12;}\\{ - 8}&a&{25;}&{}&{}&{}\end{array}} \right);\\ > > {\rm{p}} = {\rm{poly}}\left( {\rm{A}} \right);\\ > > {\rm{eig}} = {\rm{eign}}\left( {\rm{A}} \right)\end{array}\]

So, the characteristic polynomial and eigenvalues of *A* is shown below:

\[a\] | Characteristic polynomial | Eigenvalues |

\[31.8\] | \[ - .4 - 2.6t + 4{t^2} - {t^3}\] | \[3.1279,1, - .1279\] |

\[31.9\] | \[.8 - 3.8t + 4{t^2} - {t^3}\] | \[2.7042,1,.2958\] |

\[32.0\] | \[2 - 5t + 4{t^2} - {t^3}\] | \[2,1,1\] |

\[32.1\] | \[3.2 - 6.2t + 4{t^2} - {t^3}\] | \[1.5 \pm .9747i,1\] |

\[32.2\] | \[4.4 - 7.4t + 4{t^2} - {t^3}\] | \[1.5 \pm 1.4663i,1\] |

The procedure to draw the graph of the above equation by using the graphing calculator is as follows:

Draw the graph of the function \(f\left( t \right) = - .4 - 2.6t + 4{t^2} - {t^3}\)**, \(g\left( t \right) = .8 - 3.8t + 4{t^2} - {t^3}\)**, \(k\left( t \right) = 2 - 5t + 4{t^2} - {t^3}\), \(x\left( t \right) = 3.2 - 6.2t + 4{t^2} - {t^3}\) and \(y\left( t \right) = 4.4 - 7.4t + 4{t^2} - {t^3}\)** **by using the graphing calculator as shown below:

- Open the graphing calculator. Select the “STAT PLOT” and enter the equation \( - .4 - 2.6t + 4{t^2} - {t^3}\) in the \({Y_1}\) tab.
- Select the “STAT PLOT” and enter the equation \(.8 - 3.8t + 4{t^2} - {t^3}\) in the \({Y_2}\) tab.
- Select the “STAT PLOT” and enter the equation \(2 - 5t + 4{t^2} - {t^3}\) in the \({Y_3}\) tab.
- Select the “STAT PLOT” and enter the equation \(3.2 - 6.2t + 4{t^2} - {t^3}\) in the \({Y_4}\) tab.
- Select the “STAT PLOT” and enter the equation \(4.4 - 7.4t + 4{t^2} - {t^3}\) in the \({Y_5}\) tab.
- Enter the “GRAPH” button in the graphing calculator.

Visualizations of graphs of the functions** **stated above are shown below:

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