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Q5.2-30E

Expert-verified
Found in: Page 267

### Linear Algebra and its Applications

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:

See the step by step solution

## Step 1: Determine characteristic polynomial and eigenvalues for the matrix A

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 $$p = 2 - 5t + 4{t^2} - {t^3}$$, $${\rm{eig}} = \left\{ {2,1,1} \right\}$$.

## Step 2: Determine characteristic polynomial and eigenvalues for the matrix A for each value of set a

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 $31.8$ $- .4 - 2.6t + 4{t^2} - {t^3}$ $3.1279,1, - .1279$ $31.9$ $.8 - 3.8t + 4{t^2} - {t^3}$ $2.7042,1,.2958$ $32.0$ $2 - 5t + 4{t^2} - {t^3}$ $2,1,1$ $32.1$ $3.2 - 6.2t + 4{t^2} - {t^3}$ $1.5 \pm .9747i,1$ $32.2$ $4.4 - 7.4t + 4{t^2} - {t^3}$ $1.5 \pm 1.4663i,1$

## Step 3: Plot the graph of characteristic polynomial

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:

1. 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.
2. Select the “STAT PLOT” and enter the equation $$.8 - 3.8t + 4{t^2} - {t^3}$$ in the $${Y_2}$$ tab.
3. Select the “STAT PLOT” and enter the equation $$2 - 5t + 4{t^2} - {t^3}$$ in the $${Y_3}$$ tab.
4. Select the “STAT PLOT” and enter the equation $$3.2 - 6.2t + 4{t^2} - {t^3}$$ in the $${Y_4}$$ tab.
5. Select the “STAT PLOT” and enter the equation $$4.4 - 7.4t + 4{t^2} - {t^3}$$ in the $${Y_5}$$ tab.
6. Enter the “GRAPH” button in the graphing calculator.

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