Question

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:

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 functionsstated above are shown below: