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Magazine Chapter 5: Q5.2-30E (page 267)
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.
Short Answer
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:
Step by step solution
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 \(p = 2 - 5t + 4{t^2} - {t^3}\), \({\rm{eig}} = \left\{ {2,1,1} \right\}\).
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 |
\[31.8\] | \[ - .4 - 2.6t + 4{t^2} - {t^3}\] | \[3.1279,1, - .1279\] |
\[31.9\] | \[.8 - 3.8t + 4{t^2} - {t^3}\] | \[2.7042,1,.2958\] |
\[32.0\] | \[2 - 5t + 4{t^2} - {t^3}\] | \[2,1,1\] |
\[32.1\] | \[3.2 - 6.2t + 4{t^2} - {t^3}\] | \[1.5 \pm .9747i,1\] |
\[32.2\] | \[4.4 - 7.4t + 4{t^2} - {t^3}\] | \[1.5 \pm 1.4663i,1\] |
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:
- Open the graphing calculator. Select the 鈥淪TAT PLOT鈥� and enter the equation\( - .4 - 2.6t + 4{t^2} - {t^3}\)in the\({Y_1}\)tab.
- Select the 鈥淪TAT PLOT鈥� and enter the equation\(.8 - 3.8t + 4{t^2} - {t^3}\)in the\({Y_2}\)tab.
- Select the 鈥淪TAT PLOT鈥� and enter the equation\(2 - 5t + 4{t^2} - {t^3}\)in the\({Y_3}\)tab.
- Select the 鈥淪TAT PLOT鈥� and enter the equation\(3.2 - 6.2t + 4{t^2} - {t^3}\)in the\({Y_4}\)tab.
- Select the 鈥淪TAT PLOT鈥� and enter the equation\(4.4 - 7.4t + 4{t^2} - {t^3}\)in the\({Y_5}\)tab.
- Enter the 鈥淕RAPH鈥� button in the graphing calculator.
Visualizations of graphs of the functionsstated above are shown below:
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